ARMYBR~1

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A. Structural Picture of Bridge (Superstructure & Substructure).

X

12750

1250 10250 1250

1250 450 0 0 0 0 9350 0 0 450 1250

Toe

55

00

2525

450

25

25

300

0

1775

0

BF - 01

450

0BF-03 BF-02

0

0

1250 450 3450 2450 3450 450 1250

12750

X

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Abutment Plan

375

300 700

49

47

61

47

400 450 150

H1

H

Toe

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0 0 0 450 1775 0 300 450 2525

0 0 0 450 1775 300 450 2525

5500

A1 and A2

y X

y X

0 25.000 0

0.300 24.400 0.300

25.000

PC Girder Plan

20002

00

20

0

20

00

1

3

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0

18

00 2

00

0

2 2

15

0

4 4

16

50

0 150 150 0

350

7 7

00

0 350 0

350

PC Girder Section X-X Mid Section

0

0

0

02 2

0

0 100 100 0

1

5

33

66

1

3

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0

0

0

0 0 0

PC Girder Section Y-Y End Section

75 250 75

75

17

00

16

25

3

1

2

2

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PC Cross Girder Section

1070

35

70

300200

2000

4000

40

00

24,400 Toe

Effect area = 87.11 m2

Wind Action Area

B

H1

Ka.g.H1

ka.g.H1

Girder

Slab and wearing Curb

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Earth Pressure

Sylhet 18.833 19.592 19.592 18.833 Balagonj

16.556 17.315 16.86 16.86 17.315 16.556

A1 P1 P2 A2

13.683 6.3 6.3

36.6 42.68 36.6

RL 15.080 3.716 6.200 13.657

10000

1750 150 600 5000 2500

455

11

01

5

1200

300

1750 6500 1750

8015

H2

Ka.g.H2

1

23 3

4

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8015

11

01

5

600

4007

.5

700

300

1200

11000

2100

6000

3000 5000 3000

6000

11000

Pile Cap

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1800 A1P1 P1P2 1800

1200

11

01

5

10

56

0

1200

300 300

8615 8160

10000

Tie Beam

1200

Strap Beam

1200 1200

11000Pier P1

End Span / 2 Mid Span / 2

34

00

1070 1070

37

76

12

76

9

300 300

dia circular column

Pier cap

Pile Cap

Railing

slabcurb

Railing

slabcurb

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12

76

9

34

00

200 200

37

76

1830 22061

1,0

15

12541200 1200300 300

8615

A1P1 Span P1P2 Span

1500 1500

300

1200

Pier P1

Mid Span / 2 End Span / 2

31

70

1070 1070

27

15300 300

200 200

16001145

1254

40

39

.00

0

slab

Girder

slab

Girder

Railing

slabcurb

Girder

Railing

slabcurb

Girder

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40

39

.00

0P1P2 Span P2A2 Span

1200

Pier P2

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5500

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2147

600

300

1900

1200 H2

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900 900

1800

1

4

2, 3

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300 1200 300

2100 1800 2100

6000

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12

95

7

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12

95

73

81

2

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38

12

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B. Load Analusis of Bridge Structure :

a. National Highway b. Regional Highway working stress for pile load bearing capacity calcc. District Road 3d. Bridge Type Simple Supported Single Span T- Girdere. End Span Length,C/C 24.400 mf. Total Girder Length 25.000 mg. Cross Section Category RCCh. Stem Height 1.900 mi. H1, Back wall top to Well cap top 4.947 mj. H2, Well cap height 1.200 mk. H, Back wall top to Well cap bottom (Toe) 6.147 m

0.441 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.)

7.935

2. Calculations for Dead Loads Of Super-structure :

A. Super Imposed Loads on Girders : i. Exterior Girder

ItemCross Section Height Interval Load

m m m (m) kN/m1.Railing Post 0.225 0.225 1.070 2.000 0.650

2. Railing Beam number 0.175 0.175 3.000 2.205

3Curb(a). Side walk 0.300 1.475 10.620 (b). (-)Conduit 0.925 0.225 (4.995)

4. Slab 0.200 2.125 10.200 Total for (1+2+3+4)= 18.680

5. Wearing Course 0.075 0.650 1.121

6.Conduit (For Utility) 0.925 0.225 2.081 Total for (5+6)= 3.202

Sub - total = 43.765

ii. Interior Girder Height Width Volume Loadm m m kN/m

1. Slab 0.200 2.000 0.400 9.600 - - -

- 2. Wearing Course 0.075 2.000 0.150 3.450

Sub - total = 13.050

l. Ko , Co-efficient of Active Horizontal Earth Pressure ,

m) Dp-ES, DL Surcgarge Horizontal Pressure Intensiy (ES) kN/m2

m2

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B. Self-Weight of RCC Girders & X-Girders: i. Exterior Girder

Item nosLength Width Height Volume Weight

m m m m3 kN i. Central Section Section 1 0.00 25.000 2.000 0.200 0.000 - Section 2 0.00 25 0.000 0.000 0.000 - Section 3 0.00 25 0.150 0.000 0.000 - Section 4 2.00 25 0.150 0.150 0.563 13.500 Section 5 1.00 25 0.350 1.800 15.750 378.000 Section 6 2.00 25 0.000 0.000 0.000 - Section 7 2.00 25 0.000 0.000 0.000 - Sub - total = 391.500

ii. End SectionSection 1 1.000 0 - - 0.000 - Section 2 2.000 0 - - 0.000 - Section 3 1.000 0 - - 0.000 - Sub - total = -

Exterior Girder Self weight kN = 391.50

ii. Interior Girder Interior Girder Self weight kN = 391.50

iii. Cross Girder Number of Cross Girder = 5.00 nos Length of Cross Girder = 6.60 m

Cross Girder nos Length width Height Volume Weight

m m m m3 KN Section 1 1.000 6.600 0.250 1.700 2.805 67.320 Section 2 2.000 6.600 0.075 0.075 0.037 0.891 Sub - total = 68.211

Cross Girder Self weight kN = 68.211

3. Dead Load Reaction for Abutment

Item nosLength LOAD/METER Load for 1 no. Total Load Reaction

m kN/m kN kN kN i. Exterior Girdera. Super Imposed without WC & Uti 2 25.000 18.680 467.001 934.001 467.001 b. Self Weight 2 25.000 15.660 391.500 783.000 391.500 c. Wt. from X-Girder 5 0.825 10.335 42.632 85.264 42.632 d. Total from (a+b+c) 36.045 901.133 1,802.265 901.133 e. Super Imposed only WC & Utili 2 25.000 3.202 80.062 160.125 80.062 f. Total Load from Exterior Girder 2 1,023.827 2,047.654 981.195

ii. Interior Girdera. Super Imposed without WC. 3 25.000 9.600 240.000 720.000 360.000 b. Self Weight 3 25.000 15.660 391.500 1,174.500 587.250

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c. Wt. from X-Girder 5 1.650 10.335 85.264 255.791 127.896 d. Total from (a+b+c) 28.671 716.764 2,150.291 1,075.146 e. Super Imposed only WC. 3 25.000 3.450 86.250 258.750 129.375 d. Total for Int.-Girder 3 803.014 2,409.041 1,204.521

Total Dead Load Reaction for Abutment (kN) = 2185.715625

i. Wheel load 145.000 145.000 35.000

kN kN kN

4.300 4.300 15.800 24.400

Ra = 287.111 kN Rb = 37.889 kN

287.111 kN 37.889 kN

Wheel load Reaction for 2 Lane on Each Abutment 574.221 kN

1.330 (Applicable only for Truck/Wheel Load & Tandem Loading & Tandem Loading)

763.714 kN

iii. Lane Load on Bridge Decka. Lane Load Intensity = 9.300 kN/m/Lane

b. Length of Bridge Deck = 25.000 mc.Lane Load for 1(One) Lane Bridge = 232.500 kN

Reaction on each Abutment due to Lane Load for 2 Lane Bridge Deck 232.500 kN

iv. Pedestrian Load

a. Pedestrian Load Intensity = 3.600 b. Width of Each Sidewalk = 1.250 m

c. Length of Sidewalk = 25.000 md. Pedestrian Load on 1no Sidewalk = 112.500 kN

Reaction on each Abutment due to Pedestrian Load on 2nos Sidewalk 112.500 kN

Total Live Load Reaction for Abutment 1,108.714 KN

5. Calculations for Vertical Load and Resisting Moment from Abutment Components, Soil &

ComponentsHeight Width Length Weight

m m m KN m KN-m i. Back Wall 2.147 0.300 9.350 144.536 3.225 466.129 ii. Bridge Seat-Rect.-1 0.600 1.000 9.350 134.640 2.875 387.090

Ra =Rb =

RWheel =

Dynamic Load Allowance Factor (DLAF) IM =

Wheel Load Reaction on Each Abutment including DLAF (IM) RWheel-F =

RLane =

kN/m2

RPedestrain =

RLL =

Also Resisting Moment from Superstructure Loads (Live & Dead) :

Arm from Toe

Resisting Moment

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" Rect.-2 0.300 0.450 9.350 30.294 2.750 83.309 " Tri -1 0.300 0.400 9.350 13.464 3.108 41.851 " Tri-2 0.300 0.150 9.350 5.049 2.475 12.496 iii. Stem-i 1.900 0.450 9.350 95.931 2.750 263.810 Stem-ii 1.900 0.300 9.350 127.908 3.075 393.317

551.822 1,648.001 iv.Wing Wall -Well Part-2 nos 4.947 0.450 2.975 317.894 4.013 1,275.551 Cantilever Part-1,Rect-2 no 2.000 0.450 3.000 129.600 7.000 907.200 Cantilever Part-2,Trin-2 nos 1.500 0.450 3.000 48.600 7.500 364.500

366.494 1,640.051 v.Counterfort WallAbutment- (3 nos) - - - - - - WingWall-1 (2 nos) 4.947 0.450 3.450 184.325 5.275 972.316 WingWall-2 (2 nos) - - - - - - WingWall-3 (2 nos) - - - - - -

184.325 972.316 vi. Well Cap-Part-1(Rectangular) 1.200 12.750 2.750 1,009.800 4.125 4,165.425

Part-2 (Rectangular) 1.200 7.250 2.750 574.200 1.375 789.525 Part-3 (Semi Circular) 2nos. 1.200 2.750 2.750 940.950 2.063 1,940.710

2,524.950 6,895.660 l for Substructure Components= 3,627.592 11,156.028 vii. Back Fill (BF-1) 4.947 9.350 2.525 2,102.265 4.238 8,908.347

" (BF-2) - - - - - - " (BF-3) - - - - - -

2,102.265 8,908.347 Vertical load components of Abutment (sub-structure) Total (KN)= 5,729.857 Total (KN-m) 20,064.375

KN m KN-m

vii.Dead Load Reaction & Moment from Superstructure 2,185.716 2.750 6,010.718

vii. Live Load Reaction & Moment from Superstructure 1,108.714 2.750 3,048.964

Total Vertical Load & Moment from Structure 9,024.287 29,124.057

6. Calculation of Horizontal Loads (Pressures) & Overturning Moments

i. Earth Pressure, P 995.609 KN 2.849 m 2,836.490 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.) 600.820 KN 0.600 m 360.492

72.871 KN 0.400 m 29.148 ii. Horizontal Surcharge Load on Abutment 402.334 KN 3.674 m 1,477.974 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.) 121.398 KN 0.600 m 72.839

162.500 KN 7.947 m 1,291.388 = 162.500kNActing at1.800m Above Deck ; AASHTO-LRFD-3.6.4)

27.265 KN 2.000 m 54.530 on Vertical Faces Perpendicular to Traffic. AASHTO-LRFD-3.8.1.2.3).

69.686 KN 4.000 m 278.746 on Vertical Surface of all Supperstructure Elements. AASHTO-LRFD-3.8.1.2.2; Table-3.8.1.2.2-1.)

Sub Total for i+ii+iii =

Sub Total for iv =

Sub Total for v =

Sub Total for vi =

Sub Total for vii=

PV = MR =

ii. Braking Force (25% of Truck Weight

iii. Wind Load on Substructure (0.950kN/m2

iv. Wind Load on Superstructure (0.800kN/m2

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v. Wind Load on Live Load ( 0.550kN/m Acting a 13.750 KN 7.947 m 109.271 1.800 m Above Deck ; AASHTO-LRFD-3.8.1.3)

2,466.234 KN 6,510.878

KN-m

i. Factor of Safety Against Overturning Mr / Mo 4.473 OK, FS more than 2

ii. Factor of Safety Against Sliding 0.4 * Pv / Vh 1.464 Not OK

iii. Location of Resultant Force,Lr (Mr - Mo) / Pv 2.506 m

Total Horizontal Forces VH = (KN)Total Overturning Moment, MO =(KN-m)

7. Calculations for Factor of Safety Against Overturning & Factor of Safety Against Sliding Abutment Structure Against Imposed Vertical Loads/Forces & Horizontal Forces/Pressure on Abutment & Also Location of Rsultant Forces in Y-Y Direction Well Cap Toe

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7. Calculations for Factor of Safety Against Overturning & Factor of Safety Against Sliding Abutment Structure Against Imposed Vertical Loads/Forces & Horizontal Forces/Pressure on Abutment & Also Location of Rsultant Forces in Y-Y

STRUCTURAL DESIGN OF DELPARA BRIDGE AT 18.25km ON COX'S BAZAR-TEKNUF MARIN DRIVE ROAD UNDER COX'S BAZAR ROAD DIVISION (IMPLEMENTION AUTHORITY ;- 16 ECB BANGLADESH ARMY).

C. Checking of Stability for Abutment & Well Cap Against Loads of Different Components & Applied Forces on Structural Elements :

1

a) Type of Sub-soil : a) At Borehole No-BH07 (Cox's Bazar End), from GL (GL is - 2.25m from Road Top Level) up to 2.50m depth Sub-soil posses Loss gray fine Silty sand having SPT Value ranging 7 to 12. Whereas in next 0.75m from depth 2.50m to 3.75m there exists Medium dense gray fine sand with SPT value ranging from 12 to 40. From depth about 3.75m there exists Bed-rock (Gray Shale) having 50 and over SPT values .b) At Borehole No-BH08 (Teknuf End), from GL (GL is - 2.25m from Road Top Level) upto 2.75m depth Sub-soil posses Medium dency gray sandy silt having SPT Value renging 12 to 37. In next 2.15m (About depth 2.75m to 4.80m) there exists Medum densey gray fine sand with SPT value renging from 37 to 50.From depth about 4.80m there exists Bed rock (Gray Shale) having SPT value 50 over.

b) Type of Foundation : Due to its Geographical position, Marin Drive Road have every risk to effected by Wave action & Cyclonic Strom from Sea. More over the Slain Water is also an important factor for RCC Construction Works in these area. In Designing of any Permanent Bridge/Structure on this Road, specially in Foundation Design all the prevalling adverse situations should be considered for their Survival andDurability. Though as per Soil Investigation Report there exist Loss to Mediumdency gray sandy silt on Seashore Sub-soil, but due to ground their formation those posses a very poor Mechanical bonding among it contitutent.But there exites Bed-rock at a considerably short depth (About 3.75m to 4.80m) from the Ground Level.Presence of Bed-rock is an important for the Foundation of any Structue on this Road. To encounter all mentioned adverse situations Provisionof RCC Caissons embedded into the Bed-rock will be best one as Foundation of Bridges on this Road. RCC Caissons embedded into the Bed-rock will be aSolid mass to save guard the Structure against Errosion, Sliding, Overturning etc. which caused by the Wave action & Cyclonic Strom. More over against Salinity effect necessary meassary can provide for RCC Caissions. Thus it isrecommended to Provide RCC Caissions embedded into the Bed-rock at least1.50m into Bed-rock as Foundation of Delpara Bridge.

c) Type of Abutment : Wall Type Abutment.

d) Type of Wing-walls : Wall Type Wing Walls Integrated with Abutment Wall having Counterforts over Well & Cantilever Wings beyond Well.

e) Design Criteria : Ultimate Stress Design (AASHTO-LRFD-2004).

2 Design Data in Respect of Unit Weight, Strength of Materials, Soil Pressure & Multiplier Factors :

Description Notation Dimensions Unit.

Information about Soil, Foundation, Abutment & Wing-walls:


i)

9.807

a) Unit weight of Normal Concrete 2,447.23

b) Unit weight of Wearing Course 2,345.26

c) Unit weight of Normal Water 1,019.68

d) Unit weight of Saline Water 1,045.17

e) Unit weight of Earth (Compected Clay/Sand/Silt) 1,835.42

ii)

a) Unit weight of Normal Concrete 24.00

b) Unit weight of Wearing Course 23.00

c) Unit weight of Normal Water 10.00

d) Unit weight of Saline Water 10.25

e) Unit weight of Earth (Compected Clay/Sand/Silt) 18.00

iii) Strength Data related to Ultimate Strength Design( USD & AASHTO-LRFD-2004) :

a) 21.00 MPa

b) 8.40 MPa

c) 23,855.62 MPa

d) 2.89

e) 2.89 MPa

f) 410.00 MPa

g) 164.00 MPa

h) 200000 MPa

iv) Permanent & Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1 ; Table 3.4.1-1&2 :

a) 8.384 Let n 8

b) r 19.524c) k 0.291d) j 0.903

e) R 1.102

v) Sub-soil Investigation Report & Side Codition Data:

a) N 50 Over

Unit Weight of Different Materials in kg/m3:

(Having value of Gravitional Acceleration, g = m/sec2)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3

Unit Weight of Different Materials in kN/m3:

wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3

Concrete Ultimate Compressive Strength, f/c (Normal Concrete) f/

c

Concrete Allowable Strength under Service Limit State (WSD) = 0.40f/c fc

Modulus of Elasticity of Concrete, Ec = 0.043gc1.50Öf/

c Ec

(AASHTO LRFD-5.4.2.4).

Poisson's Ration = 0.63Öf/c = 0.63*21^(1/2), subject to cracking and

considered to be neglected (AASHTO LRFD-5.4.2.5).

Modulus of Rupture of Concrete, fr = 0.63Öf/c Mpa fr


Steel Ultimate strength, fy (60 Grade Steel) fy

Steel Allowable Strength under Service Limit State (WSD) = 0.40fy fs

Modulus of Elasticity of Reinforcement, Es for fy = 410 MPa ES

Modular Ratio, n = Es/Ec ³ 6 =

Value of Ratio of Steel & Concrete Flexural Strength, r = fs/fc Value of k = n/(n + r) Value of j = 1 - k/3

Value of R = 0.5*(fckj)

SPT Value as per Soil Boring Test Report,


b) 33

c) Recommended Allowable Bearing Capacity of Soil as per Soil Investigation p 770

d) d 19 to 24

e) AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.

0.34 to 0.45v)

0.395

vi) Permanent & Dead Load Multiplier Factors Under Strength Limit State (USD) :

a) 1.250 Applicable to All Components Except Wearing Course & Utilities (Max. value of Table 3.4.1-2)

b) 1.500 (Max. value of Table 3.4.1-2)

c) Multiplier Factor for Horizontal Active Earth Pressure on Substructure 1.500

value of Table 3.4.1-2)

d) Multiplier Factor for Vertical Earth Pressure on Substructure Components of 1.350

e) Multiplier Factor for Surchage Pressure on Substructure Components of 1.500

(Max. value of Table 3.4.1-2)

vii) Live Load Multiplier Factors :a) Multiplier Factor for Multiple Presence of Live Load ( No of Lane = 2)-m m 1.000

(ASSHTO LRFD-3.6.1.1.1)

b) 1.750

c) IM 1.330 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1;(Applicable only for Truck Loading & Tandem Loading)

d) 1.750

e) 1.750

f) 1.750

Corrected SPT Value for N>15, N/ = 15 + 1/2(N - 15) = 15 + 1/2(50 - 15) N/

= 15 + 1/2(50 - 15) = 32.5 . Say N/ = 33

kN/m2

Report witht SPT Value 50 over, p = 7.2 Ton/ft2. = 770kN/m2

For Back Filling with Clean fine sand, Silty or clayey fine to medium sand O

Angle of Friction with Concrete surface, d = 190 to 240,

Recommended Co-efficient of Lateral Active Earth Pressure Ka-recom Ka-recom

= 0.34 to 0.45 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.)Provided Co-efficient of Active Earth Pressure is Average of, Ka-recom Ka

Dead Load Multiplier Factor for Structural Components & Attachments-DC gDC

Dead Load Multiplier Factor for Wearing Course & Utilities-DW, gDW

gEH

Components of Bridge-EH; Applicable to Abutment & Wing Walls, (Max.

gEV

Bridge-EV; Applicable toAbutment & Wing Walls, (Max. value of Table 3.4.1-2)

gES

Bridge-ES; Horizontal & Vertical Loads on Abutment & Wing Walls,

Multiplier Factor for Truck Loading (HS20 only)-LL-Truck. gLL-Truck

Multiplier Factor for Vhecular Dynamic Load Allowence-IM as per Provision of

Multiplier Factor for Lane Loading-LL-Lane gLL-Lane

Multiplier Factor for Pedestrian Loading-PL. gLL-PL.

Multiplier Factor for Vehicular Centrifugal Force-CE gLL-CE.


g) 1.750

h) 1.750

i) 1.000

j) STRENGTH - III 1.400

l) STRENGTH - V 1.000

k) 1.000

l) 1.000 (With Elastomeric Bearing).

m) 1.000 (With Elastomeric Bearing).

n) 1.000 (With Elastomeric Bearing).

o) 1.000 (With Elastomeric Bearing).

p) 1.000 (With Elastomeric Bearing).

q) -

r) -

t) 1.000

vii) Permanent & Dead Load Multiplier Factors for Service Limit State Design (WSD) According to AASHTO-LRFD-3.4.1 ; Table 3.4.1-1&2 :






Multiplier Factor for Vhecular Breaking Force-BR. gLL-BR.

Multiplier Factor for Live Load Surcharge-LS gLL-LS.

Multiplier Factor for Water Load & Stream Pressure-WA gLL-WA.

Multiplier Factor for Wind Load on Structure-WS gLL-WS.

Multiplier Factor for Wind Load on Live Load-WL gLL-WL

Multiplier Factor for Water Load & Stream Pressure-FR gLL-FR.

Multiplier Factor for deformation due to Uniform Temperature Change -TU gLL-TU.

Multiplier Factor for deformation due to Creep on Concrete-CR gLL-CR.

Multiplier Factor for deformation due to Shrinkage of Concrete-SH gLL-SH.

Multiplier Factor for Temperature Gradient-TG gLL-TG.

Multiplier Factor for Settlement of Concrete-SE gLL-SE.

Multiplier Factor for Earthquake -EQ gLL-EQ.

Multiplier Factor for Vehicular Collision Force-CT gLL-CT.

Multiplier Factor for Vessel Collision Force-CV gLL-CV.



gEH


gEV




ii) Live Load Multiplier Factors for Service Limit State Design (WSD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :

a) Multiplier Factor for Multiple Presence of Live Load ( No of Lane = 2)-m m 1.000 (ASSHTO LRFD-3.6.1.1.1)

b) 1.000

c) IM 1.300 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1 ; SERVICE - II(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000

f) SERVICE - II 1.300

g) SERVICE - II 1.300

h) 1.000

i) 1.000

j) SERVICE - IV 0.700

l) SERVICE - II 1.300

k) 1.000






gES



















q) -

r) -

t) 1.000

3

4 Dimension of Different Sub-Structural Components & RCC Well for Foundation:

i)

a) Height of Abutment Wall from Bottom of Well Cap up to Top of Back Wall, H 6.147 m

b) Height of Abutment Wall from Top of Well Cap up to Top of Back Wall, H1 4.947 m

c) Height of Abutment Well Cap, 1.200 m





Sketch Diagram of Abutment & Wing wall:

C

C

Dimensions of Sub-Structure.

hWell-Cap.

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

5225

1200

2000

6350

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

10250

9350

12750

3450 3450 3450600 600 600600

34503450

60 0

2750

60 0

2525

1775

5500

3000

450

450

450

450

2150

2150

2450

2750

RL-5.00m

RL-2.20m

3000 2100

300


d) Height of Abutment Steam 1.900 m

e) Depth of Girder including Deck Slab 2.000 m

f) Height of Bearing Seat 0.147 m

g) 2.147 m

h) Height of Wing Wall 4.947 m

i) Length of Wing Walls upon Well cap 2.975 m

j) Width (Longitudinal Length) of Abutment Well Cap, 5.500 m

Length (Transverse Length) of Abutment Well Cap, 12.750 mk)

10.250 ml) Direction.

m) Inner Distance in between Wing Walls (Transverse), 9.350 m

n) Thickness of Abutment Wall (Stem) at Bottom 0.750 m

o) Thickness of Abutment Wall (Stem) at Top 0.450 m

p) Thickness of Counterfort Wall (For Wing Wall) 0.450 m

q) Number of Wing-Wall Counterforts (on each side) 1.000 No's

r) Clear Spacing between Counerfort & Abutment Wall at Bottom 1.775 m

s) Average Spacing between Counerfort & Abutment Wall 2.375 m

t) 2.825 m

u) Thickness of Wing Walls within Well Cap, 0.450 m

v) Thickness of Cantilever Wing Walls 0.450 m

w) Length of Cantilever Wing Walls 3.000 m

x) Height of Rectangular Portion of Cantilever Wing Walls 2.000 m

hSteam.

hGir.

hBearing

Height of Back Wall = hGir. + hBearing hb-wall

H-W-Wall

LW-W-Well-Cap

WAb-Cap

LAb-T-W-Cap

Transverse Length of Abutment Wall (Outer Face to Outer Face) in X-X LAB-Trans.

LWW-T-Inner

t.-Ab-wal-Bot.

t.-Ab-wal-Top.

tWW-Countf.

NW-W-count

SClear-Count& Ab-Bot.

SAver-Count&Ab.

= (tAB-Wall-Bot + tAb-Wall-Top)/2+SClear-Count& Ab-Bot.

Effective Span of Wing Wall Counterfort = SAver-Count + tWW-Countf SEfft-Count.

t-Wing-wall

tw-wall-Cant.

Lw-wall-Cant.

hw-wall-Cant.-Rec.


y) Height of Triangular Portion of Cantilever Wing Walls 1.500 m

z) Longitudinal Length of Well Cap on Toe Side from Abutment Wall 2.525 mOuter Face.

z-i) Average Length (Longitudinal) of Well Cap on Heel Side from 2.825 m

ii) Dimensions of RCC Well for Foundation.

a) 5.500 m

b) 12.750 m

c) Depth of Well from Bottom of Well Cap up to Bottom of Well Curb 6.325 m

d) Wall thickness of Well, 0.600 m

f) Thickness of Partition Walls of Well, 0.600 m

g) Diameter of Outer Circle, 5.500 m

h) 4.300 m

i) 7.250 m

j) 4.300 m

k) Number of Pockets within Well 3.000 Nos

l) 4.300 m(Longitudinal Span Length).

m) 3.450 m(Transverse Span Length).

n) 3.450 m(Transverse Span Length).

o) 2.750 m

p) 63.633

q) Total Length of Staining of Well (Main & Partitions) through Center line 38.494 m

hw-wall-Cant.-Tri.

L-W-Cap-Toe.

L-W-Cap-Heel-Aver.

Abutment Wall Face.= SAver.-Count.& Ab. + tWW-Count.

Width of Well in Y-Y Direction (In Longitudinal Direction) WWell-Y-Y

Length of Well in X-X Direction (In Transverse Direction) LWell-X-X

HWell-pro.

tWall.

tWall-Perti

DOuter.

Diameter of Inner Circle = DOuter - 2* tWall DInner.

Transverse Length of Rectangular Portion of Well Cap =LWell-X-X - DOuter LRect.

Length of Partition Walls = DOuter - 2*tWall LParti.

NPock.

Distance between Inner Faces of Pockets in Y-Y Direction SPock-Y-Y.

Distance between Inner Faces of Outer Pockets in X-X Direction SPock-X-X-Outer.

Distance between Inner Faces of Central Pocket in X-X Direction SPocket-X-X-Central.

Width of Well from its c.g. Line in X-X. = WWell-Y-Y/2 W1/2-Well-Y-Y

Surface Area of Well at Top & Bottom Level = pDOuter2/4 + LRect*DOuter AWell. m2

LStaining.

= p*(DOuter+ DInner)/2 + 2*LRect. + 2*LParti


r) 66.879

s) 2.939 m

t) 2.100 m

u) (4.750) m

5 Calculations for Safe Bearing Capacity (S.B.C.) of Soil for Well (Caisson) as Abutment Foundation .

a) h 6.850 m

b) 2.283 m

c) 6.325 m

d) Calculated Soil Bearing Capacity at Bottom Level of Well Foundation 164.811

e) Since the Soil Bearing Capacity as per Soil Investigation Report,

6 Checking for Stability of Well Cap as Abutment Base against all applied Forces:

i) Imposed Vertical Loads/Forces upon Abutment Well Cap (As per Design Calculation Sheet - C) :

a) Dead Load Reaction from Super-Structure 2,185.716 kN

b) Live Load Reaction from Super-Structure (Pedestrians, Wheel & Lane Load) 1,108.714 kN

c) Dead Load Reaction from Sub-Structure & Earth Pressure, 5,729.857 kN

Surface Area of Well Cap = LAb-T-W-Cap*W1/2-Well-Y-Y + 0.5*pDOuter2/4 AWell-Cap. m2

+ LRect*W1/2-Well-Y-Y

Distance of c.g. (X-X) Line from Well Cap Toe Face bc.g.-Y-Y.

= (LAb-T-W-Cap*(W1/2-Well-Y-Y)2*1.50+ (0.5*pDOuter2/4)*0.50*DOuter*3/4

+ LRect*W1/2-Well-Y-Y2/2)/AWell-Cap.

RL of Highest Flood Level (HFL) HFLRL

RL of Maximum Scoring Level (MSL) MSLRL

Height between RL of HFL & RL of MSL = HFLRL - MSLRL

x = (2.100- (-)4.75)m

Minimum Depth required for Bottom Level of Well from MSL= h/3 HWell-req

Provided Depth from Well Cap Bottom up to Bottom Level of Well H-Well-pro.

pCal kN/m2

pCal = 3.5(N-3)*{(B+0.3)/2B}*a*b + W; Where N = 50 over, the Field SPT value;

N/ = 33,the Corrected SPT value; f = 36.900, the Angle of Shearing Resistance of Soil; B =5.000m,Width of Well for Foundation; D = 6.250m, Depth of Well;a = 0.50, for Submerge of Well Bottom; b = (1+D/5B) > 1.20 = (1+6.25/5*5) = 1.2

and W = Soil Pressure per m2 at Bottom Level of Well = D*g = 6.250*18.00kN/m2

= 112.500kN/m2

Thus the Calculated Soil Bearing Capacity pCal = 3.5(N-3)*{(B+0.3)/2B}*a*b + W,

= (3.5*(50-3)*((5.00+0.3)/(2*5))*0.50*1.20 + 112.500)kN/m2 = 164.811kN/m2

Since the Soil Bearing Capacity as per Soil Investigation Report, p = 770kN/m2 > pCal

= 164.811kN/m2, thusthe Well Foundation is OK in respect of S.B.C.

RDL-Supr.

RLL-Supr.

RDL-Sub.


d) Total Vertical Forces due to Dead & Live Load 9,024.287 kN(Super-Structure + Sub-Structure)

e) Moment due to Dead Load Reaction from Super-Structure 6,010.718 kN-m

f) Moment due to Live Load Reaction from Super-Structure 3,048.964 kN-m

g) Moment due to Dead Load & Soil Pressure for Sub-Structure 20,064.375 kN-m

h) Total Resisting Moment due to Vertical Forces (Dead & Live Load from 29,124.057 kN-m Super-Structure + Sub-Structure, Earth Pressure)

i) Total Horizontal Forces due to Earth Pressure, Surcharge, Braking, Wind- 2,466.234 kNLoad, Dead Load Friction etc.

j) Total Overturning Moments due to Horizontal Forces 6,510.878 kN-m

ii) Checking Against Overturning.

a) 4.473 OK

b)

iii) Checking Against Sliding.

a) 1.464 Not OK

b)

iv) Calculation of Eccentricity in respect of c.g. Line of Pile Cap in X-X Direction due to Applied Loads &Moments ( Vertical & Horizontal):

a) 22,613.179 kN-m

b) x 2.506 m

c) 2.939 m

d) e 0.433 m

e) 1/6th 0.490 m

f)Pile Cap in Y-Y direction, Factor of Safety against Overturning & Sliding are within limit range, Thus the Structure is a Stable One in all respect.

PV

MDL-Supr.

MLL-Supr.

MDL-Sub.

MR

PH

MO

Factor of Safety against Overturning, FSOverturn = MR/MO ³ 2. FSOverturn

Since FSO > 2 , thus the Structure is safe in respect of Overturning.

Factor of Safety against Sliding, FSSlid = 0.4*PV / PH ³ 1.50. FSSlid

Since FSSlid > 1.50, thus the Structure is safe in respect of Sliding.

Net Moment or Algebraic sum of Moment about 'B' = MR - MO MN

Distance of Resultant Forces from Well Cap Toe Face, x = MN /PV


Eccentricity, e = bc.g.-Y-Y. - x

1/6th Distance of bc.g.-Y-Y. from c.g. towards Well Wall Cap Toe = bc.g.-Y-Y./6

Since the Calculated Eccentricity has (+) ve value & its Location is within Middle 1/3rd Portion of the


7 Checking for Stability of Well Cap as Abutment Base Without Superstrucre Loads (DL & LL) :

i) Applied Loads Moments :

a) Dead Load Reaction from Sub-Structure & Earth Pressure, 5,729.857 kN

b) Moment due to Dead Load & Soil Pressure for Sub-Structure 20,064.375 kN-m

c) Total Horizontal Forces due to Earth Pressure, Surcharge & Wind Load on 2,220.297 kNSubstructure.

d) Total Overturning Moments due to Horizontal Forces 4,831.474 kN-m

ii) Checking Against Overturning.

a) 4.153

b)

iii) Checking Against Sliding.

a) 1.032

b)make the Structure a Safe one in respect of Sliding.

iv) Calculation of Eccentricity in respect of c.g. Line of Pile Cap in X-X Direction due to Applied Loads & Moments ( Vertical & Horizontal):

a) 14,334.518 kN-m

b) x 2.502 m

c) 2.939 m

d) e 0.437 m

e) 1/6th 0.490 m

f)Cap in Y-Y direction, Factor of Safety against Overturning is within limit range, though Safety Factor against Sliding is less than limit range but the Well Cap is Integrated with RCC Well Structure, thus the Structure is a Stable One in all respect without Superstructure Loads also.

RDL-Sub.

MDL-Sub.

PH

MO

Factor of Safety against Overturning, FSOverturn = MR/MO ³ 2. FSOverturn

Since FSOvertur > 2 , thus the Structure is safe in respect of Overturning.

Factor of Safety against Sliding, FSSlid = 0.4*PV / PH ³ 1.50. FSSlid

Though FSSlid < 1.50, but the Well Cap would be a Integrated Component of the RCC Well, which will

Net Moment or Algebraic sum of Moment about 'B' = MR - MO MN

Distance of Resultant Forces from Well Cap Toe Face, x = MN /PV


Eccentricity, e = bc.g.-Y-Y. - x

1/6th Distance of bc.g.-Y-Y. from c.g. towards Well Wall Cap Toe = bc.g.-Y-Y./6

The Calculated Eccentricity has (+) ve value & its Location is within Middle 1/3rd Portion of the Well

D.

1 General Data for Construction Materials of Different Structural Components :


i)

9.807






ii)







a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa

iv) Strength Data related to Working Stress Design & Service Load Condition ( WSD & AASHTO-SLS ) :

a) 8.384 Say n 8

b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102

Design Data, Factors & Methods for Analysis of Flexural Design of Structural Elements:



gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


c



c Ec


Poisson's Ration = 0.63Öf/c = 0.63*21^(1/2), subject to cracking and considered

to be neglected (AASHTO LRFD-5.4.2.5).

Modulus of Rupture of Concrete, fr = 0.63Öf/c = 0.63*21^(1/2)Mpa fr





Modular Ratio, n = Es/Ec ³ 6 =



v) Design Data for Resistance Factors for Conventional Construction (AASHTO LRFD-5.5.4.2.1). :

a) For Flexural & Tension in Reinforced Concrete 0.90

b) For Flexural & Tension in Prestressed Concrete 1.00

c) For Shear & Torsion of Normal Concrete 0.90

d) For Axil Comression with Spirals or Ties & Seismic Zones at Extreme 0.75 Limit State (Zone 3 & 4).

e) For Bearing on Concrete 0.70

f) For Compression in Strut-and-Tie Modeis 0.70

g) For Compression in Anchorage Zones with Normal Concrete 0.80

h) For Tension in Steel in Anchorage Zones 1.00

i) For resistance during Pile Driving 1.00

j) For Partially Prestressed Components in Flexural with or without Tension 1.00

PPR

410.00

vi)

a) 0.85

Acc =Area of Concrete Element in Compression of Crresponding Strength.

b) 0.85

vii) Ultimate Strength Data for Design of Prestressing Components ( USD & AASHTO-LRFD-2004) :

a)

b) Tensial Strength for Strand with Grade 250 having Diameter 6.35 to 15.24mm, 1,725 Mpa

c) Tensial Strength for Strand with Grade 270 having Diameter 9.37 to 15.24mm, 1,860 Mpa

(Respective Resistance Factors are mentioned as f )

fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

Flexural value of f of Compression Member will Increase Linearly as the Factored Axil Load Resitance,

fPn, Decreases from 0.10f/cAg to 0.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.

fFlx-PPR.

Resistance Factor f = 0.90 + 0.10*(PPR) in which,

PPR = Apsfpy/(Apsfpy + Asfy), where; PPR is Partial Prestress Retio.

As = Steel Area of Nonprestressing Tinsion Reinforcement in mm2 As mm2

Aps = Steel Area of Prestressing Steel mm2 Aps mm2

fy = Yeiled Strength of Nonprestressing Bar in MPa. fy N/mm2

fpy = Yeiled Strength of Prestressing Steel in MPa. fpy N/mm2

b Factors for Conventional RCC & Prestressed Concrete Design (AASHTO LRFD-5.7.2.2). :

Flexural value of b1, the Factor of Compression Block in Reinforced Concrete b1

up to 28 MPa.i) For Further increases of Strength of Concrete after 28 MPa agaunst each 7MPa

the value of b1 will decrese by 0.05 & the Minimum Value of b1 will be 0.65.

ii) For Composite Concrete Structure, b1avg = Σ(f/cAccb1)/Σ(f/

cAcc); where,

Value of b for Flexural Tension of Reinforcement in Concrete b

For Uncoated & Stress-relieved 7 (Seven) Wire according to AASHTO-LRFD Bridge ConstructionSpecifications (AASHTO-LRFD-5.4.4) will be; i) AASHTO M 203/M 203M (ASTM A 416/A 416M), orii) AASHTO M 275/M 275M (ASTM A 722/A 722M).

fpu-250-Str.

fpu-270-Str.

d) Tensial Strength for Type-1 Plain Bar having Diameter 19 to 35mm, 1,035 Mpa

e) Tensial Strength for Type-2 Deformed Bar having Diameter 16 to 35mm, 1,035 Mpa

f) Yield Strength for Strand with Grade 250 having Diameter 6.35 to 15.24mm, 1,466 Mpa

g) Yield Strength for Strand with Grade 270 having Diameter 9.37 to 15.24mm, 1,581 Mpa

h) Yield Strength for Type-1 Plain Bar having Diameter 19 to 35mm, 880 Mpa

i) Yield Strength for Type-2 Deformed Bar having Diameter 16 to 35mm, 828 Mpa

j) Modulus of Elastacity for Strand 197,000 MPa

j) Modulus of Elastacity for Bar 207,000 MPa

2 Different Load Multiplying Fatcors for Strength Limit State Design (USD) & Load Combination :

i) Formula for Load Factors & Selection of Load Combination :

a)

Here:

For Strength Limit State;

1.000

1.000

1.000

ii) Permanent & Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1 ; Table 3.4.1-1&2 :

a) 1.250 Applicable to All Components Except Wearing Course & Utilities (Max. value

fpu-Ty-1-P-Bar

fpu-Ty-1-P-Bar

fpy-250-Str.

= 85% of Tensial Strength (fpu).

fpy-270-Str.


fpy-Ty-1-P-Bar


fpy-Ty-1-P-Bar


Ep-Strandy

Ep-Bar

Formula for Load Factors Q = Σ ηigiQi £ f Rn = Rr; (ASSHTO LRFD-1.3.2.1-1 & 3.4.1-1)

Where, ηi is Load Modifier having values

ηi = ηDηRηI ³ 0.95 in which for Loads a Maximum value of gi Applicable; (ASSHTO LRFD-1.3.2.1-2), &

ηi = 1/(ηDηRηI) £ 1.00 in which for Loads a Minimum value of gi Allpicable; (ASSHTO LRFD-1.3.2.1-3)

gi = Load Factor; a statistically based multiplier Applied to Force Effect, f = Resistance Factor; a statistically based multiplier Applied to Nominal Resitance,

ηi = Load Modifier; a Factor related to Ductility, Redundancy and Operational Functions,

ηi = ηD = 1.00 for Conventional Design related to Ductility, ηD

ηi = ηR = 1.00 for Conventional Levels of Redundancy , ηR

ηi = ηI = 1.00 for Typical Bridges related to Operational Functions, ηl

Qi = Force Effect,

Rn = Nominal Resitance,

Ri = Factored Resitance = fRn.


of Table 3.4.1-2)







iii) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000

l) 1.000


gEH


gEV


gES














(With Elastomeric Bearing).





q) -

r) -

t) 1.000

3 Different Load Multiplying Fatcors for Service Limit State Design (WSD) & Load Combination :

i) Permanent & Dead Load Multiplier Factors for Service Limit State Design (WSD) According to AASHTO-LRFD-3.4.1 ; Table 3.4.1-1&2 :



















gEH


gEV


gES


b) 1.000

c) IM 1.000 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1 (SERVICE - I);(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

3 Intensity of Different Imposed Loads (DL & LL) & Load Coefficients :




















i) Coefficient for Lateral Earth Pressure (EH) :

a) 0.441

b) f 34.000

c) Angle of Friction with Concrete surface & Soli d 19 to 24

AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.

d) 0.34 to 0.45 dim

ii) Dead Load Surcharge Lateral/Horizontal Pressure Intensity (ES); AASHTO-LRFD-3.11.6.1. :

a) Constant Horizontal Earth Pressur due to Uniform Surcharge, 0.007935

7.935

b) 0.441 Earth Pressure,

c) 0.018

iii) Live Load Surcharge Vertical & Horizontal Pressure Intensity (LS); AASHTO-LRFD-3.11.6.4. :

a) Constant Earth Pressur both Vertical & Horizontal for Live Load 0.007141

Surcharge on Abutment Wall (Perpendicular to Traffic), Where; 7.141

0.004761

4.761

b) Constant Horizontal Earth Pressur due to Live Load Surcharge for 0.008331

Wing Walls (Parallel to Traffic), Where; 8.331

0.004761

4.761

c) k 0.441

d) 1835.424

e)

f) g 9.807

g) 900.000 mm

600.000 mm

Coefficient of Active Horizontal Earth Pressure, ko = (1-sinff ) ,Where; ko

f is Effective Friction Angle of Soil


Effective Friction Angle of Soil, f = 340 .(Table 12.9, Page-138, RAINA,s Book)

O

Value of Tan d (dim) for Coefficient of Friction. Tan d = 0.34 to 0.45 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.)

Dp-ES N/mm2

Dp-ES = ksqs in Mpa. Where; kN/m2

ks is Coefficien of Earth Pressure due to Surcharge = ko for Active ks

qs is Uniform Surcharge applied to upper surface of Active Earth Wedge(Mpa) qS N/mm2

= wE*10-3N/mm2

Dp-LL-Ab<6.00m N/mm2

kN/m2

Dp-LS = kgsgheq*10-9 Dp-LL-Ab³6.00m N/mm2

kN/m2

Dp-LL-WW<6.00m N/mm2

kN/m2

Dp-LS = kgsgheq*10-9 , Dp-LL-WW³6.00m N/mm2

kN/m2

ks is Coefficien of Latreal Earth Pressure = ko for Active Earth Pressure.

gs is Unit Weight of Soil (kg/m3) gs kg/m3

Since wE, the Unit Wieght of Soil = 18kN/m3, thus gs = wE*10^3/g

g is Gravitational Acceleration (m/sec2), AASHTO-LRFD-3.6.1.2. m/sec2

heq is Equivalent of Height of Abutment Wall Soil for Vehicular Load (mm). heq-Ab<6.00m.

Having, H < 6000mm & for having H ³ 6000mm ; heq-Ab³6.00m.

AASHTO-LRFD-3.11.6.4; Table-3.11.6.4-1.

h) Width of Live Load Surcharge Pressure for Abutment having 900.000 mm 0.900 m

AASHTO-LRFD-3.11.6.4; Table-3.11.6.4-1. 600.000 mm 0.600 m

i) 1050.000 mm

600.000 mm AASHTO-LRFD-3.11.6.4; Table-3.11.6.4-2.

j) Width of Live Load Surcharge Pressure for Wing Walls, 600.000 mm 0.600 m

600.000 mm 0.600 m

iv) Wind Load Intensity on Superstructure Elements (WS) :

a) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.000800 Mpa

Elements in Lateral Direction of Wind Flow (Parallel to Traffic). 0.800 AASHTO-LRFD-3.8.1.2.2; Table-3.8.1.2.2-1.

b) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.0009000 Mpa

Elements in Longitudinal Direction of Wind Flow (Perpendicular to Traffic). 0.900

v) Wind Load Intensity on Substructure Elements (WS) :

a) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.000950 Mpa

0.950

b) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.001645 Mpa

Elements in Longitudinal Direction (Perpendicular to Traffic). 1.645

(AASHTO-LRFD-3.8.1.2.3).

vi) Wind Load Intensity on Live Load (WL) :

a) Horizontal Wind Load Intensity on Live Load upon Superstructure 0.550 N/mmin Longitudinal Direction (Parallel to Traffic). = 0.550 N/mm, having 0.550 kN/m

b) Horizontal Wind Load Intensity on Live Load upon Superstructure 0.500 N/mmin Lateral Direction (Perpendicular to Traffic) = 0.500N/mm having 0.500 kN/m

Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.

Weq-Ab³6.00m.

heq is Equivalent of Height of Abutment Wall Soil for Vehicular heq-WW<6.00m.

Load (mm). Having, H < 6000mm & for having H ³ 6000mm ; heq-WW³6.00m.

Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

pWind-Sup-Let.

kN/m2

pWind-Sup-Long.

kN/m2

pWind-Sub-Let.

Elements in Lateral Direction (Parallel to Traffic). = 0.0019*cos600 Mpa, kN/m2

Considering 600 Skew Angle of Main Force; (AASHTO-LRFD-3.8.1.2.3).

pWind-Sub-Long.

kN/m2

= 0.0019*sin600 Mpa; Considering 600 Skew Angle of Main Force;

pWind-LL-Sup-Let.

action at 1800mm above Deck & Considering 600 Skew Angle of Force;for Two Lane Bridge. (AASHTO-LRFD-3.8.1.3; Table- 3.8.1.3-1).

pWind-LL-Sup-Long.

action at 1800mm above Deck & Considering 600 Skew Angle of Main Force; for Two Lane Bridge.(AASHTO-LRFD-3.8.1.3; Table- 3.8.1.3-1).

vii) Intensity on Breaking Force (BR) :

a) Intensity of Horizontal Breaking on Superstructure is the Greater value of 162.500 kN i) 25% of the Axle Weight of Design Truck/Design Tendem, or 162.500 kN ii) 5% of Design (Truck + Lane Load) or (Design Tendem + Lane Load) 55.750 kN Breaking Force is for Two Lane Bridge & its Action at 1800mm above Deck.(AASHTO-LRFD-3.6.4).

4 Calculations of Effects due to Imposed Deformations under Strut-and-Tie Model, (AASHTO-LRFD-5.6.3) :

i) Structural Modeling (AASHTO-LRFD-5.6.3.2) :

a)

i) For Unreinforced Compressive Struts 179928.000 N

ii) For Reinforced Compressive Struts 2488118.954 N

b)

i) For Unreinforced Compressive Struts 257040.000 N

ii) For Reinforced Compressive Struts 3554455.649 N

c) f 0.70

ii) Proportioning of Strength for Unreinforced Compressive Struts, (AASHTO-LRFD-5.6.3.3.1) :

a) 257040.000 N257.040 kN

b) 17.850

0.002 mm/mm, the PricapalTensile Strain of Crack Concrete due to Factored Load. Thus

18.357 17.850

c) 14,400.000

of Strut is 6-times the Main bar Diameter & Width is Width of Component.

32 mm & Width of Girder

450

iii) Proportioning of Strength for Reinforced Compressive Struts, (AASHTO-LRFD-5.6.3.3.2) :

a) 3,554,455.649 N

3,554.456 kN

10 804.25

8,042.48

iv) Proportioning of Tension Ties, (AASHTO-LRFD-5.6.3.4.) :

pLL-Sup-Break.

pLL-25%-Truck.

pLL-5%-(Tru+Lane.)

Factored Resistance of Strut-and-Tie, Pr = fPn .

Pr-Unrin.

Pr-Rin.

Pn = Nominal Resistance of Strut or Tie in N.

Pn-Unrin.

Pn-Rin.

f = Tension/Compression Resistance Factor as required for the Component.

Nominal Resistance of Unreinforced Strut in N, Pn-Unrin. = fcuAcs; where, Pn-Unrin.

fcu = Limiting Compressive Stress in MPa as per AASHTO-LRFD-5.6.3.3.3. fcu. N/mm2

= f/c/(0.8+172e l) £ 0.85f/

c, here, e l =

f/c/(0.8+172e l) = & 0.85f/

c =

Acs = Effective X-Sectional Area of Strut in mm2 under the provision of Acs. mm2

AASHTO-LRFD-5.6.3.3.2. For Strut Anchored by Reinforcement the Length

For RCC Girder Diameter of Bar fBar =

b = mm. Acs = fBar*b

Nominal Resistance of Reinforced Strut in N, Pn-Rin. = fcuAcs + fyAss ; where, Pn-Rin.

Ass is Area of Reinforcement of Strut in mm2 . For RCC Girder nos. of Bars

Nbar = nos.X-Area of bar,Af =p*fBar2/4 = mm2

Thus Ass = Af*Nbar.= mm2

a) 5,605,606.604 N 5,605.607 kN

b) 13,672.211

17

c) 0.000

d) 0.000

v) Proportioning of Node Regions at Bearing Positions on Support, (AASHTO-LRFD-5.6.3.5.) :

a) The Compressive Stress of Node Regions Bounded by Compressive Struts

b) The Compressive Stress of Node Regions Anchoring by a One-direction

c) The Compressive Stress of Node Regions Anchoring by a Two-direction

vi) Design AASHTO HS20 Truck Loading :

a) Axle to Axle distance 1.800 m

b) Wheel to Wheel distance 4.300 m

c) Rear Wheel axle Load (Two Wheels) 145.000 kN

d) Rear Single Wheel Load 72.500 kN

e) Middle Wheel axle Load (Two Wheels) 145.000 kN

f) Middle Single Wheel Load 72.500 kN

g) Front Wheel axle Load (Two Wheels) 35.000 kN

h) Front Single Wheel Load 17.500 kN

vii) Design AASHTO Lane Loading :

a) Design Lane Loading is an Uniformly Distributed Load having Magnitude of 9.300 N/mm 9.300N/mm through the Length of Bridge for 1 (One) Lane of Bridge & acting 9.300 kN/m over a 3.000m Wide Dcak Strip in Transverse Direction. Thus Lane Load per meter Length of Bridge for 1 (One) Lane = (9.300*1000/1000)kN/m

b) Design Lane Loading is an Uniformly Distributed Load having Magnitude of 3.100 kN/m/m-Wd.

9.300N/mm through the Length of Gridge for Single and acting over a 3.000m 0.003100 N/mm/mm-Wd.

Wide Strip in Transverse Direction. Thus Intensity of Lane Load per meter Length & for per meter Width = 9.300/3.000kN/m/m-Wd.

viii) Design AASHTO Pedestrian Loading :

Nominal Resistance of Tension Tie in N, Pn-Ten. = fyAst + Aps(fpe + fy); where, Pn-Tie.

Ast is Total Area of Longitudanal Reinforcement of Tie in mm2 . For RCC Girder Ast. mm2

Total nos.Bar Nbar = nos. Thus, Ast = Af*Nbar.

Aps is Area of Prestressing Steel in mm2 . For RCC Girder, Aps = 0. Aps. mm2

fpe is Stress in Prestressing Steel after loss in MPa. For RCC Girder, fpe = 0 fpe N/mm2

fc-Node-Bearing N/mm2

and Brearing Area will be, fc-Node-Bearing £ 0.85fÖf/c

fc-Node--1-Dir-Ten-Tie N/mm2

Tension Ties will be, fc-Node-1-Dir-Ten-Tie £ 0.75fÖf/c

fc-Node-2-Dir-Ten-Tie N/mm2

Tension Ties will be, fc-Node-2-Dir-Ten-Tie £ 0.65fÖf/c

DAxel.

DWheel.

LLRW-Load

LLRS-Load

LLMW-Load

LLMS-Load

LLFW-Load

LLFS-Load

LLLane

LLLane-Int.

a) Design Pedestrian Loading is an Uniformly Distributed Load having Magnitude 0.003600

3.600 the total Wide of Sidewalk.

ix) Design Live Line Loading of Sidewalk :

a) 14.600 kN/mSlab Span not Exceeding 1800 mm from Center of Exterior Girder & where Exists a Structurally Continuous Concrete Railing, there will be a Longitudinal

Action of Line Load will be at Distance 300mm from the Face of Railing & itwould Replace the Outside Row of Wheel Load. Since the Design Strip is inTransverse Direction with 1.000m Width, thus Action on Line Load will like a Concentrated Load having Magnitude = 14.600*1000/1000kN.

4 Calculations of Respective Events for Design of Flexural And Axial Force Effects, (AASHTO-LRFD-5.6.3) :

i) Stress in Prestressing Steel at Nominal Flexural Resistance for Flexural Members, (AASHTO-LRFD-5.7.3.1):

a) In Components having Bonded Prestressing Tendon in

In which,

k

ii) For T-Section behavior of Components, mm

iii) For Rectangular Section behavior of Components, mm

Where,

b mm

mm

mm

mm

c mm

0.85

ii) Factored Flexural Resistance for Prestressed or RCC Elements (AASHTO-LRFD-5.7.3.2.1):

LL-Pedest N/mm2

of 3.600*10-3MPa through the Length of Sidewalk on both side and acting over kN/m2

According to AASHTO-LRFD-3.6.1.3.4 Bridges having Overhanging Deck PLL-Line.

& Uniformly Distributed Line Live Load (LL) of Magnitude of 14.600N/mm.

fps N/mm2

One Axis, the Average Stress in Prestressing Steel fps = fpu(1-k*c/dp) ³ 0.5* fpu

i) k = 2.00*(1.04-fps/fpu)

cT-Secton

the value of c = (Apsfps+Asfy -A/f/y -0.85b1f/c(b-bw)hf)/(0.85f/

cb1bw+kApsfpu/dp)

cRec-Secton

the value of c = (Apsfps+Asfy -A/f/y)/(0.85f/cb1b+kApsfpu/dp)

iv) As = Steel Area of Nonprestressing Tinsion Reinforcement in mm2 As mm2

v) Aps = Steel Area of Prestressing Steel mm2 Aps mm2

vi) A/s = Steel Area of Nonprestressing Compression Reinforcement in mm2 A/

s mm2

vii) fy = Yeiled Strength of Nonprestressing Tension Bar in MPa. fy N/mm2

viii) fy = Yeiled Strength of Nonprestressing Tension Bar in MPa. f/y N/mm2

ix) fpy = Yeiled Strength of Prestressing Steel in MPa. fpy N/mm2

x) b = Width of Compression Flange in mm.

xi) bw = Width of Web in mm. bw

xii) hf = Height of Compression Flange in mm. hf

xiii) dp = Distance of Extreme Compression Fiber from Prestressing Tendon Centroid dp

in mm.xix) c = Distance of Neutral Axis from Compression Face of Component in mm.

xx) b1 = Compression Stress Block Factor in mm. b1

a) N-mm

N-mm

f 0.90

b)

Where;

b mm

mm

mm

mm

mm

mm

0.85

a mm

c)

d)

5 Limits For Maximum & Minimum Reinforcement, (AASHTO-LRFD-5.7.3.3) :

i) Limits For Maximum Reinforcement, (AASHTO-LRFD-5.7.3.3.1) :.

a) With Maximum Amount of Prestressed & Nonprestressed Reinforcement for a

Factored Flexural Resistance for any Section of Component, Mr = fMn, where; Mr

i) Mn is Nominal Resistance Moment for the Section in N-mm Mn

ii) f is Resistance Factor for Flexural in Tension of Reinforcement/Prestressing.

The Nominal Resistance for a Flanged Section subject to One Axis Stress as per AASHTO-LRFD-5.7.3.2.2 is

Mn = Apsfps(dp-a/2) + Asfy(ds-a/2) - A/sf/

y(d/s-a/2) + 0.85f/

c(b-bw)b1hf(a/2-hf/2).

i) As = Steel Area of Nonprestressing Tinsion Reinforcement in mm2 As mm2

ii) Aps = Area of Prestressing Steel in mm2 Aps mm2

iii) A/s = Steel Area of Nonprestressing Compression Reinforcement in mm2 A/

s mm2

iv) fy = Yeiled Strength of Nonprestressing Tension Bar in MPa. fy N/mm2

vi) f/y = Yeiled Strength of Nonprestressing Tension Bar in MPa. f/

y N/mm2

vii) fps = Average Strength of Prestressing Steel in MPa. fps N/mm2

viii) b = Width of Compression Flange in mm.

ix) bw = Width of Web in mm. bw

x) hf = Depth of Compression Flange for I or T Member in mm. hf

xi) dp = Distance of Extreme Compression Fiber from Prestressing Tendon dp

Centroid in mm.

xii) ds = Distance of Centroid of Nonprestressed Tensial Reinforcement from the ds

Extreme Compression Fiber in mm.

xiii) d/s = Distance of Centroid of Nonprestressed Compression Reinforcement d/

s

from the Extreme Compression Fiber in mm.

xix) b1 = Compression Stress Block Factor in mm. b1

xx) a = cb1; Depth of Equivalent Stress Block in mm.

For Nonprestressing Element of Structure the corresponding values against Aps,fps,dp all are = 0,. Whereas for

Singly Reinforced Structural Elements the Respective values of f/y & d/

s = 0. Thus Equation for Nominal Resistance

of I or T Member with having Flenge & Web Components Stand at Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

For Nonprestressing Element of Structure having only Rectangular Component b = bw & hf = 0. Thus Equation for

Nominal Resistance of Singly Reinforced Structural Member Stand at Mn = Asfy(ds-a/2)

c/de

Section c/de £ 0.42 in which;

i) de = (Apsfpsdp + Asfyds)/(Apsfps + Asfy), where ;

ii) As = Steel Area of Nonprestressing Tinsion Reinforcement in mm2 As mm2

mm

mm

b) For a Structure having only Nonprestressed Tensial Reinforcement the values of

ii) Limits For Manimum Reinforcement, (AASHTO-LRFD-5.7.3.3.2) :

a) For Section of a Flexural Component having Prestressed & Nonprestressed Tensile Reinforcements or only with

b) N-mmwhere;

at Extreme Fiber where Tensile Stress is caused by Externally Applied Forces

N-mm

2.887

c) N-mm

d) N-mm

e) N-mm

f) N-mm

6 Control of Cracking By Distribution of Reinforcement, (AASHTO-LRFD-5.7.3.4) :

iii) Aps = Area of Prestressing Steel in mm2 Aps mm2


vi) fps = Average Strength of Prestressing Steel in MPa. fps N/mm2


Centroid in mm.

xii) ds = Distance of Centroid of Nonprestressed Tensial Reinforcement from the ds

Extreme Compression Fiber in mm.

Aps, fps & dp are = 0. Thus Equation for value of de stands to de = Asfyds/Asfy &

thus de = ds .

Nonprestressed Tensile Reinforcements should have Minimum Resisting Moment Mr ³ 1.2*Mcr or 1.33 Times the Calculated Factored Moment for the Section Based on AASHTO-LRFD-3.4.1-Table-3.4.1-1, which one is less.

The Cracking Moment of a Section Mcr = Sc(fr + fcpe) - Mdnc(Sc/Snc - 1) £ Scfr Mcr

i) fcpe = Compressive Stress in Concrete due to effective Prestress Forces only fcpe N/mm2

after allowance for all Prestressing Losses in MPa.

ii) Mdnc = Total Unfactored Dead Load Moment acting on the Monolithic or Mdnc

Noncomposite Section in N-mm.

iii) Sc = Section Modulus for the Extreme Fiber of the Composite Section where Sc mm3

Tensile Stress Caused by Externally Applied Loads in mm3.

iv) Snc = Section Modulus of Extreme Fiber of the Monolithic or Noncomposite Snc mm3

Section where Tensile Stress Caused by Externally Applied Loads in mm3.

v) fr = Modulus of Rupture of Concrete in RCC in Mpa,(AASHTO LRFD-5.4.2.6). fr N/mm2

For Nonprestressing & Monolithic or Noncomposite Beam or Elements, Mcr

Sc = Snc & fcpe = 0, thus Equation for Cracking Moment Stands to Mcr = Sncfr

Thus Calculated value of Mcr according to respective values of Equation Mcr-1

The value of Mcr = Scfr Mcr-2

Computed value of Mcr = 1.33*MExt Factored Moment due to External Forces Mcr-3

a) The Tensial Reinforcement of all Concrete Elements under Service Limit State

mm Tension Bar considering the Max. Clear Cover = 50mm.

A Dividing the Total Concrete Area bounded in between Extreme Tension Face & a Straight Line parallel to Neutral Axis of Component having equal distance from the Centrioed of Main Tension Reinforcement Bars on both sideded by the total Number of Main Bars as Tensial Reinforcement (The Max. Clear Cover = 50mm.)

Max. value of Crack Width Parameter for Moderate Exposure Components 30,000 N/mm

Max. value of Crack Width Parameter for Severe Exposure Components 23,000 N/mm

Max. value of Crack Width Parameter for Buried Components 17,000 N/mm

iv) In Transverse Design of Segmental Concrete Box-Girder the Max. value of 23,000 N/mm

the full Nominal Concrete Strength.

N/mm

bd mm

b)for the Locations beyond Decompression State & the Prestressing Steel are in Crack Section.

c)Limit State Design, Flexural Tension Reinforcement also should be Provided on Flange for the Width of lesser value as Mentioned;

d)between Bearings, Additional Longitudinal Reinforcement should Provide on ExcessWidth Portion of Flange. These Additional Longitudinal Reinforcement should be

e)

fsa N/mm2

Load according to AASHTO-LRFD-3.4.1-Table-3.4.1-1 (Except Deck Slab Design

under AASHTO-LRFD-9.7.2), fsa =Z/(dcA)1/3 £ 0.6fy, Where;

i) dc = Depth of Concrete Extrime Tension Face from the Center of the Closest dc

For a Particular Section of the Component dc = fBar/2 + CCover-Bot. (Max. 50mm)

ii) A = Area of Concrete Surrounding a Single Tension Bar, which is Calculted by mm2

Thus, A = 2*(CCover-Bot.(Max.= 50mm) + (NB-Layers+ sBar-Vert)/2)/NBar

iii) Z = Crack Width Parameter for Cast In Place Components in N/mm. Other then Cast in Place Box Culvert;

ZMode-Expo.

ZSever-Expo.

ZBuried.

ZBox-Gir-Trans.

Crack Parameter, Z £ 23000N/mm for any Loads Applied prior to attainment of

v) In Cast in Place RCC Box Culvert the Crack Parameter Z £ 27500/b , ZBox-Culvert

in which; b = (1+dc/0.7d), where, d = Distance from Compression Face to Centriod of Tension Reinforcement. Thus value of 27500/b 27500/b

In Bonded Prestressed Concrete, Equation fsa =Z/(dcA)1/3 £ 0.6fy, is applicable

If Tension prevailes in Flanges of RCC T-Girder & Box Girder, under Service

i) On the Effective Flange Width as according to AASHTO-LRFD-4.6.2.6

ii) On the Width Equal to 1/10 of Average Length of Adjacent Spans between Bearings.

For Effective Flange Width Greater than 1/10 of Average Length of Adjacent Spans

At least 0.40 % of the X-Sectional Area of the Excess Width Portion.

If the value of dc of Nonprestrssed or Partially Prestressed Concrete Member Exceeds 900 mm, than Uniformly

i) Required Steel Area of Skein Reinforcement,

v) The Max. Spacing of Longitudinal Skein Reinforcement should be less than

Distributed Longitudinal Skein Reinforcement would be required on both Side Faces of the Component over a

Distance of dc/2 from the nearest Flexural Tension Reinforcement.

Ask mm2

Ask ³ 0.0001(dc - 760) £ As +Aps) ; where,

ii) Aps = Area of Prestressing Steel in mm2 Aps mm2

iii) As = Area of Tensile Reinforcement As mm2

iv) Total Area Longitudinal Skein Reinforcement should be <= 1/3rd of (Aps + As)

either dc/6 or 300 mm.

N/mm/mm-Wd.

E. Design of Deck Slab for RCC Girder Bridge :

1

10.250

1.475 1.475 7.300

1.070 1.250 1.250

0.3000.200

150x150

0.950 1.650 1.650 1.650 1.650 0.950

0.350

1.125 2.000 2.000 2.000 2.000 1.125

2 Structural Data :

i) Type of Bridge Structure : Single Span Simple Supported RCC Girder Bridge.

ii) Type of Deck Slab : RCC Deck Slab having Over Hanging Cantilever on both Side.

iii) Type of Side Walk & Railings: RCC Side Walk with RCC Pedestrian Type Railing & Railing Posts.

iv) Dimensions of Superstructure Components of Bridge Deck & Girder:


a) Span Length of Bridge (C/C Distance between Bearings) 24.400 m

b) Carriageway width of Bridge Deck, 7.300 m.

c) Width of Side Walk on both of Bridge Deck 1.250 m.

d) Height of Railing Post 1.070 m

Sketch of Bridge Deck Section:

Strength

WC-way

WSidewalk.

HR-Post.

75 75

22 5

175*175

e) Longitudinal Width of Railing Post 0.255 m

f) Transverse Width of Railing Post 0.255 m

g) Interval Railing Post. 2.000 m

h) Number of Rails on Each Side Walk. 3.000 nos.

i) Vertical Depth of Railing 0.175 m

j) Horizontal Width of Railings 0.175 m

k) Height of Curb/Wheel-Guard/Riling Post Base 0.300 m

l) Width of Curb/Wheel-Guard & Riling Post Base 0.275 m

m) Total width of Bridge Deck Slab, 10.250 m.

n) Thickness of Bridge Deck Slab, 0.200 m.

o) Thickness of Wearing Course 0.075 m.

q) Thickness of Side Walk Slab, 0.075 m.

r) No's of Girders 5.000 nos.

k) C/C Distance between Main Girders 2.000 m.

l) Width of Main Girder 0.350 m.

m) Depth of Main Girder including Slab Thickness 2.000 m

n) Length of Cantilever Slab from Exterior Girder Center 1.125 m.

o) Clear Distance in between Two Main Girder Faces, 1.650 m.

p) Distance of Cantilever Slab Edge from Exterior Girder Face, 0.950 m

q) Depth of Main Girder Fillet 0.150 m

r) Width of Main Girder Fillet 0.150 m.

t) Distance between Outer Edge of Deck & Inner Edge of Curb 1.475 m

u) 0.925 m

WPost-Long.

WPost-Trans..

Int.Post

NRails.

hRailing.

bRailing.

hCurb/Guard

bCurb/Guard

WDeck

tSlab

tWC

tSlab-SW

NGir

LC/C-Gir.

bGir-Main

hGir-Main

LExt.-Gir-to-Cant.

LGir.-Face.

LCant-Ext.

hFillet.

bFillet.

WEdge-to-Curb.

Transverse Width of Utility/Fluid Space = WEdge-to-Curb. -2*bCurb/Guard WFluid.

v) 0.225 m

w) Distance of Wheel Guard Inner Edge up to Inner Face of Exterior Girder 0.175 m = WEdge-to-Curb. - LExt.-Gir-to-Cant. - bGir-Main/2

x) Distance of Wheel Guard Inner Edge up to Outer Face of Exterior 0.525 m

y) Distance of Wheel Guard Outer Edge up to Outer Face of Exterior Girder 0.250 m

v) Computation of Effective Flange Width for T-Girder RCC Bridge :

a)

= 6.100 m

2.750 m

= 2.000 m

b) Since Average Spacing of Adjacent Beams/Girders is the Least one,

2.000 m

3 Design Criterion, Type of Loading & Design Related Data :

i) Design Criterion :

a) AASHTO Load Resistance Factor Design (LRFD).

b) Type of Loads : Combined Application of AASHTO HS20 Truck Loading & Lane Loading.

ii)

9.807





e) Unit weight of Earth (Compacted Clay/Sand/Silt) 1,835.424

iii)





Depth of Utility/Fluid Space = hCurb/Guard - tSlab-SW hFluid.

LCurb-inner-Ext.

LCurb-Outer-Ext.

Girder = LCurb-inner-Ext.. + bGir-Main

LCurb-Outer.

= bGir.-Main - (bCurb/Guard - LCurb.-inner)

Under Provision of AASHTO LRFD-4.6.2.6 (4.6.2.6.1) the Flange Width of T-Girder will be the least Dimension of :

i) One-quarter of Effective Span Length = 1/4*SLength

ii) 12.0 times average Depth of Slab + Greater Thickness of Web = 12*tSlab + bGi =iii) One-half the Width of Girder Top Flange (It is not req. as there is no Addl. Top Flange)

iv) The average Spacing of Adjacent Beams/Girders = LC/C-Gir.

thus the Flange Width of Interior Girders, bFla-Gir = 2.000m bFl-Gir.


(Having value of Gravitational Acceleration, g m/sec2)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3

Unit Weight of Materials in kN/m3 Related to Design Forces :

wC kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

e) Unit weight of Earth (Compacted Clay/Sand/Silt) 18.000

iv) Strength Data related to Ultimate Strength Design( USD & AASHTO-LRFD-2004) :

a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200,000.000 MPa





d) For Axil Compression with Spirals or Ties & Seismic Zones at Extreme 0.750 Limit State (Zone 3 & 4).


f) For Compression in Strut-and-Tie Modes 0.700




j) 0.850 (AASHTO LRFD-5.7.2..2.)

k) 0.850

vi) Other Design Related Data :

a) Velocity of Wind Load in Normal Condition 90.000 km/hr

b) Velocity of Wind Load in Special Condition 260.000 km/hr

c) Velocity of Water/Stream Current Causing Water/Stream Load 4.200 m/s

5 Factors Applicable for Design of Different Structural Components under Strength Limit State (USD):


a)

wE kN/m3


c



c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).








(Respective Resistance Factors are mentioned as f or b value)

fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.

Value of b1 for Flexural Compression in Reinforced Concrete b1


VWL-Nor.

VWL-Spe.

VWA

Formula for Load Factors Q = Σ ηigiQi £ f Rn = Rr; (AASHTO LRFD-1.3.2.1-1 & 3.4.1-1)


Here:


1.000

1.000

1.000

ii) Selection of Load Multiplying Factors for Strength Limit State Design (USD) & Load Combination :

a) The Bridge will have to face Cyclonic Storms with very high Intensity of Wind Load (Wind Velocity = 260km/hr),

iii) Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :






e) Multiplier Factor for Surcharge Pressure on Substructure Components of 1.500


iv) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :

ηi = ηDηRηI ³ 0.95 in which for Loads a Maximum value of gi Applicable; (AASHTO LRFD-1.3.2.1-2), &

ηi = 1/(ηDηRηI) £ 1.00 in which for Loads a Minimum value of gi Applicable; (AASHTO LRFD-1.3.2.1-3)

gi = Load Factor; a statistically based multiplier Applied to Force Effect, f = Resistance Factor; a statistically based multiplier Applied to Nominal Resistance,





Qi = Force Effect,

Rn = Nominal Resistance,

Ri = Factored Resistance = fRn.

but those would be occasional. Thus the respective Multiplier Factors of Limit State STRENGTH I (Bridge used by Normal Vehicle without wind load) for normal operation, Limit State of STRENGTH-III (Wind Velocity exceeding90km/hr) for wind load during cyclonic storm condition and Limit State of STRENGTH-IV (Having Wind Velocity of 90 km/hr) for normal wind load only are selected as CRITICAL conditions for bridge structure.



gEH


gEV

Bridge-EV; Applicable to Abutment & Wing Walls, (Max. value of Table 3.4.1-2)

gES


a) Multiplier Factor for Multiple Presence of Live Load ( No of Lane = 2)-m m 1.000 (AASHTO LRFD-3.6.1.1.1)

b) 1.750

c) Multiplier Factor for Vehicular Dynamic Load Allowance-IM as per Provision o IM 1.330 AASHTO LRFD-3.6.2.1, Table 3.6.2.1-1;(Applicable only for Truck Loading & Tandem Loading)

d) 1.750

e) 1.750

f) 1.750

g) Multiplier Factor for Vehicular Breaking Force-BR. 1.750

h) 1.750

i) 1.000


l) Multiplier Factor for Wind Load on Live Load-WL STRENGTH - V 1.000

k) 1.000






q) -

r) -

t) 1.000





gLL-BR.




gLL-WL










6 Different Load Multiplying Factors for Service Limit State Design (WSD) & Load Combination :











b) 1.000

c) IM 1.000 AASHTO LRFD-3.6.2.1, Table 3.6.2.1-1; SERVICE - I(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000

i) 1.000



gEH


gEV


gES



Multiplier Factor for Vehicular Dynamic Load Allowance-IM as per Provision of




Multiplier Factor for Vehicular Breaking Force-BR. gLL-BR.





k) 1.000






q) -

r) -

t) 1.000

7 Philosophy in Flexural Design of Bridge Deck Slab :

a)Loading, Tandem Loadings, Pedestrian Loads etc. for whom the Bridge Structure are being build. In addition Dead

Walk Slab) including Utilities if there be any.

b) Since Casting of Deck Slab RCC Bridge would be Monolithic with closely placed Longitudinal Main Girders, thus the Deck is absolutely a One-way Continuous Slab with Overhanging Cantilever Parts on both Sides. In addition of

Walk, Conduit with Utility, Wheel Guard & Curb etc. More over the Cantilevers will also carry the Pedestrian Live

Central Portion of Deck.

b)& Moments, the Deck Slab is to be considered as Constituents of Transverse Stripes having 1.000m width.

c) For the Purpose of Design, Computations of Shearing Forces & Moments will be done in Four Different Sections.












The Bridge Deck is that Structural Component which directly carries the Live Loads (LL) of Vehicular Traffic, Lane

Loads (DL) due to Self Weight & Deck Furniture's (Riling & Riling Posts, Wheel Guard & Curb, Medians & Side

Self Weight Dead Load (DL) the Cantilevers would also carry the Dead Loads (DL) from Railing & Railing Post, Side

Loads (LL). Self Weight & Imposed Wearing Course are the Dead Load (DL) Constituents of Central Deck Portion.The Main Live Loads (LL) for the Central Portion is either Concentrated Wheel Loads of Design Truck or Similar butof Different Magnitude Tandem Loads. A Uniformly Distributed Lane Load also Constituent of Live Load (LL) for the

For the Design Purpose & Calculation of Imposed Loads (DL & LL) and Computation of respective Shearing Force

i) Section 1-1 on Outer Face of Exterior Girder is for Design of Cantilever Portion. Whereas, ii) Section 2-2 on Inner

for Design of Interior Slab Spans.

d) Sketch Diagram Showing the Sections for Design Calculations :

150x150

0.950 1.650 1.650

0.350

1.125 2.000 2.000

8

i) Effective Span Length for RCC Girder Bridge (AASHTO-LRFD-9.7.2.3):

a) For Slabs Monolithic with RCC Girders, the Effective Span Length of Deck 1.650 mSlab in between the Interior Girders is the Face to Face Distance. Thus Span Length of Interior Deck Slab = LC/C-Gir. - bGir-Main

b) For Slabs Monolithic with RCC Girders, Effective Span Length of Cantilever 0.950 mDeck Slab is the Distance from Face of Exterior Girder up to Edge of theCantilever. Thus Span Length of Cantilever Deck Slab;

ii) Effective Span Length for PC or Steel Girder Bridge (AASHTO-LRFD-9.7.2.3):

a) For Slab supported on Concrete (PC) or Steel Girder, the Effective Span mLength of Interior Deck Slab is the Distance between Flange Tips Plus the Flange Overhanging Length taken as Distance from Extreme Flange Tip tothe Face of Web, disregarding any Fillets.

b) Slab supported on Concrete PC Girder or Steel Girder, the Effective Span m

Face of Exterior Girder; iii) Section 3-3 on Middle in between Two Girders & iv) Section 4-4 of Face of Interior Girder

Selection of Effective Span Length for Deck Slab & Other Dimensions under AASHTO-LRFD-2004 :

SInter.

SCant.

= LExt.-Gir-to-Cant. - bGir-Main/2

SInter.

SCant.

1 32

1 2 3

4

4

1475

1250

1030

30 0

200

75 75

22 5

Length of Cantilever Deck Slab is the Distance between Web Face up to the Cantilever Edge, disregarding any Fillets.

iii) Required Slab Thickness for RCC Bridge Deck Slab According to Provisions of AASHTO-LRFD-2004 :

a) 155.000 mm<165

For the Case of Main Reinforcement Parallel to Traffic.

b) 175.000 mmshould not be Less than 175 mm.

c) The Provided Depth/Thickness for the Simple Supported T-Girder Bridge 200.000 mm

tslab-pro.>tslab-req. OK.

iv) Provision of Clear Cover for Deck Slab :

a)& Bottom Surface of Slab should be 75 mm from Face of main Reinforcement. Since the Top Surface of Deck Slabwill have 75 mm Thick Wearing Course, thus on this Surface may Provide with 38 mm Clear Cover. Accordingly by

9

i) Computation of Different Dead Loads (DL) Applicable for the Interior & Cantilever Span of Deck Slab :

a) For Computation of Loads Let Consider Strip in Transverse Direction having 1.000 mWidth in Longitudinal = 1.000m

b) Intensity of Uniformly Distributed Dead Load (DL) due to Self Weight of 4.800

c) 1.725

d) 2.205 as Concentrated Load (Computed from Load Analysis Sheet).

e) 0.650 Post as Concentrated Load (Computed from Load Analysis Sheet).

f) 1.980

g) 1.980

According to AASHTO-LRFD-2.5.2.6.3;Table-2.5.2.6.3-1 for a Continuous tSlab-req-1.

Span, the Thickness of Deck Slab should be = (S + 3000)/30 ³ 165mm

Here S is Calculated Effective Length of Interior Span = 1550mm

Since the Main Reinforcement is Perpendicular Traffic & tSlab-req-1 < 165m, thus the case is not Applicable.

According to AASHTO-LRFD-9.7.1.1 Depth/Thickness of a Concrete Deck tSlab-req-2.

tSlab-pro.

= 200mm which is Greater than tSlab-req-2.

According to AASHTO-LRFD-5.12.3; Table.5.12.3-1 for Structures on Costal Area , the Clear Covers both on Top

using Ante-Chlorine Admixtures for Concrete, on Bottom Surface 25 mm Clear cover may Provide.

Selection/Computation of Applied Loads/Forces (DL & LL) for Flexural Design of Deck Slab:

bStrip.

WDL-Self. kN/m2/m

Slab for per Meter Span Length & 1.000 m Wide Strip = wC*tSlab*bStrip

Intensity of Uniformly Distributed Dead Load (DL) due to Wearing Course WDL-WC. kN/m2/m

for per Meter Span Length & 1.000 m Wide Strip = wWC*tWC*bStrip.

Dead Load (DL) on Cantilever Edge on 1.000m Wide Strip From Railings WDL-Railing. kN.

Dead Load (DL) on Cantilever Edge on 1.000m Wide Strip From Railing WDL-Rail-Post. kN.

Dead Load (DL) on Cantilever Edge on 1.000m Wide Strip From Railing WDL-Post-Base. kN.

Post Base as Concentrated Load = wC*hCurb/Guard*bCurb/Guard./bStrim

Dead Load (DL) on Cantilever Part on 1.000m Wide Strip From Curb WDL-Curb. kN.

h) 1.800

i) 2.081 kNfor Utility Load as Concentrated Load (Computed from Load Analysis Sheet).

ii) Computation of Different Live Loads (LL) Applicable for the Interior & Cantilever Span of Deck Slab :

a) Since the Design Strip is 1.000m Wide over which on any Span only One 72.500 kNWheel of Design Truck or Tandem can take Position due to their Longitudinal & Transverse Spacing, thus One Rear or Middle Wheel of Design Truck having the Highest Magnitude of Loading should be Considered as Wheel

b) 3.100 Magnitude of 9.300 N/mm in Longitudinal Direction with Action in TransverseDirection over 3.000m Width & the Design Strip is in Transverse Direction of

c) 14.600 kN/mSlab Span not Exceeding 1800 mm from Center of Exterior Girder & where Exists a Structurally Continuous Concrete Railing, there will be a Longitudinal

Action of Line Load will be at Distance 300mm from the Face of Railing & itwould Replace the Outside Row of Wheel Load. Since the Design Strip is inTransverse Direction with 1.000m Width, thus Action on Line Load will like a Concentrated Load having Magnitude = 14.600*1000/1000kN.

d) 3.600

for the Cases where Sidewalk Width is Greater than 600mm. Since Provided Sidewalk Width is 1.250m & the Design Strip is in Transverse Direction with

10 Computation of Factored Loads (DL & LL) in respect of Flexural Design of Deck Slab under Provisions forStrength Limit State of Design (USD) & Service Limit State of Design (WSD) :

i) Computation of Factored Dead Loads (DL) Applicable for the Strength Limit State of Design (USD) under Provisions of AASHTO-LRFD-3.4.1 :

a) 6.000

as Concentrated Load = wC*hCurb/Guard*bCurb/Guard.

Dead Load (DL) on Cantilever Part on 1.000m Wide Strip From Side Walk WDL-S-Walk. kN/m2/m

Slab as Uniformly Distributed Load = wC*tSlab-SW*bStrip

Dead Load (DL) on Cantilever Part on 1.000m Wide Strip From Fluid WDL-Utility.

PLL-Truck.

Live Load (LL). (AASHTO-LRFD-3.6.1.2.2)

Since the Design Lane Load is a Uniformly Distributed Live Load (LL) having PLL-Lane. kN/m2

1.000m Width. Thus Magnitude of Lane Load over the 1.000 m2 area of Deck

= ((9.300*1000)/3.000)/1000 kN/m2 (AASHTO-LRFD-3.6.1.2.4)

According to AASHTO-LRFD-3.6.1.3.4 Bridges having Overhanging Deck PLL-Line.

& Uniformly Distributed Line Live Load (LL) of Magnitude of 14.600N/mm.

According to AASHTO-LRFD-3.6.1.6 on the Sidewalk Surface a Uniformly PLL-Pedst. kN/m2

Distributed Pedestrian Live Load (LL) will act having Intensity 3.6*10-3 Mpa

1.000m Width, thus Pedestrian Live Load per m2 of Sidewalk Surface

= (3.600/10^3)*10^6/10^3kN/m2.

Intensity of Factored Uniformly Distributed Dead Load (DL) due to Self WDL-Self.-USD kN/m2/m

Weight of Slab = gDC*WDL-Self

b) 2.588

c) 2.756

d) 0.813

e) 2.475

f) 2.475

g) 2.250

h) for Utility 3.122 kN

ii) Computation of Factored Dead Loads (DL) Applicable for the Service Limit State of Design (WSD) under Provisions of AASHTO-LRFD-3.4.1 :

a) 4.800

b) 1.725

c) 2.205

d) 0.650

e) 1.980

f) 1.980

g) 1.800

Intensity of Factored Uniformly Distributed Dead Load (DL) due to WDL-WC.-USD kN/m2/m

Wearing Course = gDW*WDL-WC

Factored Dead Load (DL) on Cantilever Edge From Railings as WDL-Railing.-USD kN.

Concentrated Load = gDC*WDL-Railing

Factored Dead Load (DL) on Cantilever Edge From Railing Post as WDL-Rail-Post.-USD kN.

Concentrated Load = gDC*WDL-Rail-Post.

Factored Dead Load (DL) on Cantilever Edge From Railing Post WDL-Post-Base.-USD kN.

Base as Concentrated Load = gDC*WDL-Postp-Base.

Factored Dead Load (DL) on Cantilever Part From Curb as Concentrated WDL-Curb.-USD kN.

Load = gDC*WDL-Curb.

Factored Dead Load (DL) on Cantilever Part From Side Walk Slab WDL-S-Walk.-USD kN/m2/m

as Uniformly Distributed Load = gDC*WDL-S-Walk.

Factored Dead Load (DL) on Cantilever Part From Fluid WDL-Utility.-USD

Load as Concentrated Load = gDW*WDL-Utility.

Intensity of Factored Uniformly Distributed Dead Load (DL) due to Self WDL-Self.-WSD kN/m2/m

Weight of Slab = gDC*WDL-Self

Intensity of Factored Uniformly Distributed Dead Load (DL) due to WDL-WC.-WSD kN/m2/m

Wearing Course = gDW*WDL-WC

Factored Dead Load (DL) on Cantilever Edge From Railings as WDL-Railing.WSD kN.

Concentrated Load = gDC*WDL-Railing

Factored Dead Load (DL) on Cantilever Edge From Railing Post as WDL-Rail-Post.-WSD kN.

Concentrated Load = gDC*WDL-Rail-Post.

Factored Dead Load (DL) on Cantilever Edge From Railing Post WDL-Post-Base.-WSD kN.

Base as Concentrated Load = gDC*WDL-Postp-Base.

Factored Dead Load (DL) on Cantilever Part From Curb as WDL-Curb.-WSD kN.

Concentrated Load = gDC*WDL-Curb.

Factored Dead Load (DL) on Cantilever Part From Side Walk Slab WDL-S-Walk.-WSD kN/m2/m

as Uniformly Distributed Load = gDC*WDL-S-Walk.

h) for Utility 2.081 kN

iii) Computation of Factored Live Loads (LL) Applicable for the Strength Limit State of Design (USD) under Provisions of AASHTO-LRFD-3.4.1 :

a) Factored Design Truck Wheel Load Applicable for Design of Bridge 168.744 kN

b) Factored Design Lane Load Applicable for Design of Bridge Deck 5.425

c) Factored Design Line Load Applicable for Design of Bridge Deck 25.550 kN/m

d) Factored Design Pedestrian Load Applicable for Design of Bridge Deck 6.300

iv) Computation of Factored Live Loads (LL) Applicable for the Service Limit State of Design (WSD) under Provisions of AASHTO-LRFD-3.4.1 :

a) Factored Design Truck Wheel Load Applicable for Design of Bridge 72.500 kN

b) Factored Design Lane Load Applicable for Design of Bridge Deck 3.100

c) Factored Design Line Load Applicable for Design of Bridge Deck 14.600 kN/m = m*gLL-Lane*PLL-Lane.

d) Factored Design Pedestrian Load Applicable for Design of Bridge Deck 3.600

11 Adopted Methods of Analysis in Flexural Design of Deck Slab under Provisions of AASHTO-LRFD-2004 :

i) Approximate Method of Analysis for Direct Calculation of Moments under AASHTO-LRFD-4.6.2.1.8 :

a)Composite with Reinforced Concrete Slab can consider for Determination of Live Load Moments due to VehicularLoad either from Design Truck or Design Tandem.

b)

c)

Factored Dead Load (DL) on Cantilever Part From Fluid WDL-Utility.-WSD

Load as Concentrated Load = gDW*WDL-Utility.

PLL-Truck.-USD

Deck = m*gLL-Truck*IM*PLL-Truck.

PLL-Lane.-USD kN/m2/m

= m*gLL-Lane*PLL-Lane.

PLL-Line.-USD

= m*gLL-Lane*PLL-LIne.

PLL-Pedst.-USD kN/m2/m

= m*gLL-Lane*PLL-PL.

PLL-Truck.-WSD

Deck = m*gLL-Truck*IM*PLL-Truck.

PLL-Lane.-WSD kN/m2/m

= m*gLL-Lane*PLL-Lane.

PLL-Line.-WSD

PLL-Pedst.-WSD kN/m2/m

= m*gLL-Lane*PLL-PL.

According to Article 4.6.2.1.8; Live Load Force Effects for Fully & Partially Filled Grids & for Unfilled Grid Decks

Under Provision of Article 4.6.2.1.8; Deck Slab having Main Reinforcements Perpendicular to Traffic, the Momentsboth for (+) ve & (-) ve values on Interior Spans are being Expressed by the Equations-4.6.2.1.8-1 & 4.6.2.1.8-2.

According to Equ.- 4.6.2.1.8-1; For Center-to-Center Distance of Supports, L £ 3000mm the Transverse Moments

for Deck Slab; MTransverse = 1260D0.197L0.459C in N-mm/mm

d)

Here for both Equ.

e) 1.000

for the Continuous Span = 0.80 0.800

f) D 0.606

g)

in Respective Directions.

h) Considering Design Strip in Transverse Direction, Strip Width = 1000mm. 1,000 mm

i) Effective Span Length for Deck between Interior Girders = 1650 mm 1,650 mm

j) Provided Slab Thickness = 200mm 200 mm

k) Value on Moment of Inertia for Main Reinforcement Direction; 6.667E+08

0.001

l) Value on Moment of Inertia Perpendicular to Main Reinforcement Direction; 1.100E+09

1.100E-03

m) 1.590E+13 N-mm/mm1.590E+04 kN-m/m

n) 2.624E+13 N-mm/mm2.624E+04 kN-m/m

o) 29908.110 N-mm/mm = 2000mm < 3000mm thus Equ.- 4.6.2.1.8-1 is applicable for (+) ve Moment 29.908 kN-m/m

p) The (-) ve Moment value can Calculate considering Simple Span 37,385.137 N-mm/mm 37.385 kN-m/m

i-i) Factored Moments for Strength Limit State of Design (USD) under Approximate Method of Analysis :

a) Factored (+) ve Moment for Interior Span under Strength Limit State 69611.125 N-mm/mm

69.611 kN-m/m

b) Factored (-) ve Moment for Interior Span under Strength Limit State 87013.906 N-mm/mm

87.014 kN-m/m

According to Equ.4.6.2.1.8-2; For Center-to-Center Distance of Supports, L > 3000mm the Transverse Moments

for Deck Slab; MTransverse = (5300D0.188(L1.35-20400)/L)*(C) in N-mm/mm

C is Continuity Factor having value for Simple Support Span = 1.00 & CSimple

CContinous

D is Ratio of Flexural Rigidity-DX of Deck Slab in Main Bar Direction with

Flexural Regidity - DY of Deck Slab Perpendicular to Main Bar Direction.

The Flexural Rigidity of Deck Slab in Main Bar Direction & Perpendicular to Main Bar Direction are Expressed by,

DX = EIX in N-mm2/mm & DY = EIY in N-mm2/mm where, E is Modulus of Elasticity, IX & IY are Moment of Inertia

bStrip

SInter.

tSlab.

IX mm4

IX = bStrip(tSlab)3/12 in mm4 m4

IY mm4

IX = SInter,(tSlab)3/12 in mm4 m4

Flexural Rigidity of Deck Slab in Main Bar Direction, DX = EIX DX

Flexural Rigidity of Deck Slab Perpendicular to Main Bar Direction, DY= EIY DY

Since the Center-to-Center Distance between Girders, LC/C-Gir. = L (+) MTransverse

value Calculation on Interior Span as Continuous Span having C = 0.80.

(-) MTransverse

with value of C = 0.800

(+) MTrans-USD

(USD) = mgLL-TruckIM*MTrans-UF

(-) MTrans-USD

(USD) = mgLL-TruckIM*MTrans-UF

i-ii) Factored Moments for Service Limit State of Design (WSD) under to Approximate Method of Analysis :

a) Factored (+) ve Moment for Interior Span under Service Limit State 29,908.110 N-mm/mm

29.908 kN-m/m

b) Factored (-) ve Moment for Interior Span under Service Limit State 37,385.137 N-mm/mm

37.385 kN-m/m

ii) Distribution Factor Method for Analysis of Moments for Interior Deck Spans under the Provisions of AASHTO-LRFD-4.6.2.2.2b :

a)the Live Load (Truck/Tandem) Moments for Deck Interior Spans should be based upon the Distribution Factor as

b)

c) S 1,650 mm

d) L 24,400 mm

e)

f) 200 mm

g) 1.000

h) 0.491

iii) Distribution Factor Method for Analysis of Moments for Deck Overhanging under the Provisions of AASHTO-LRFD-4.6.2.2.2d :

a)the Live Load (Truck/Tandem) Moments for Deck Overhanging Spans should be based on the Distribution Factor as

b) The Distribution Factor for Live Loads Per Lane for Moment on Deck Overhanging on face of Exterior Longitudinal

c) e 0.958

d) 525 mm

(+)MTrans-WSD

(WSD) = mgLL-TruckIM*MTrans-UF

(-) MTrans-WSD

(WSD) = mgLL-TruckIM*MTrans-WSD

According to Article-4.6.2.2.2b in Concrete Bridges having More than 4 Main Girders & Linked by Cross-Girders,

mentioned in Equation of Table- 4.6.2.2.2b-1.

The Distribution Factor for Live Loads Per Lane for Moment in Interior Beam of Multi Lane RCC T-Girder Bridge is

being Expressed by gInt.-M = 0.075+ (S/2900)0.6(S/L)0.2(0.2(Kg/Lts3))0.1; where,

S is Effective Span Length of Interior Spans for Deck's Design Strip = SInter.

L is Center-to-Center Distance between Supports of Main Girder = SBridge.

Kg is Longitudinal Stiffness Parameter in mm4 Kg mm4

ts is Thickness of Bridge Deck in mm = tSlab ts

According to Provision of Article - 4.6.2.2.2b in Preliminary Design the value Kg/Lts

of Kg/Lts = 1.000

Thus the Calculated value of Distribution Factor for Interior Span gInt.-M gInt.-M

(Having value of S = 1550 > 1100 but < 4900; ts = 200 > 110 but < 300 ;

L =24400 > 6000 but < 76000; Nb (Nos. of Girder = 5 > 4).

According to Article-4.6.2.2.2d in Concrete Bridges having More than 4 Main Girders & Linked by Cross-Girders,

mentioned in Equation of Table- 4.6.2.2.2d-1.

Girder of Multi Lane RCC T-Girder Bridge is being Expressed by gExt = egInt; where,

e is Correction Factor having value = 0.77 + de/2800; Here ,

de is a Distance in mm from Exterior Face/Web of Exterior Girder up to the de

e) 0.470

iv) Distribution Factor Method for Analysis of Shears for Interior Deck Spans According to Provisions of AASHTO-LRFD-4.6.2.2.3a :

a)Cross-Girders, the Live Load (Truck/Tandem) Shear for Deck Interior Spans should be based upon the Distribution

b)

c) S 1,650.000 mm

d) 0.635

v) Distribution Factor Method for Analysis of Shear for Deck Overhanging under the Provisions of AASHTO -LRFD-4.6.2.2.3b :

a)Cross-Girders, the Live Load (Truck/Tandem) Shear for Deck Overhanging should be based upon the Distribution

b) The Distribution Factor for Live Loads Per Lane for Shear on Deck Overhanging on face of Exterior Longitudinal

c) e 0.775

d) 525.000 mm

e) 0.492

vi) Computation of Equivalent Strip Width for Live Load Moment for Interior Deck According to Provisions of AASHTO-LRFD-4.6.2.1.3 :

Interior Edge of Curb or Traffic Barrier = WEdge-to-Curb. - SCant.

Since Position of Exterior Face/Web of Exterior Girder is within Inboard of

Slab's Outer Edge-to-Curb, thus de is of (+) ve value.

Thus Calculated Distribution Factor value for Overhanging gExt.-M = egInt-M gExt.-M

(Having value of de = 575 > (-)300 but < 1700; ts = 200 > 110 but < 300 ;

L =24400 > 6000 but < 76000; Nb (Nos. of Girder) = 5 > 4 ).

According to Article-4.6.2.2.3a in Concrete Bridges having Multi Lane with More than 4 Main Girders & Linked by

Factor as mentioned in Equation of Table- 4.6.2.2.3a-1.

The Distribution Factor for Live Loads Per Lane for Shear in Interior Beam of Multi Lane RCC T-Girder Bridge is

being Expressed by gInt.-V = 0.20 + S/3600 - (S/10700)2.00; where,

S is Effective Span Length of Interior Spans for Deck's Design Strip = SInter.

Thus the Calculated value of Distribution Factor for Interior Span gInt.-V gInt.-V

(Having value of S = 1550 > 1100 but < 4900; ts = 200 > 110 but < 300 ;


According to Article-4.6.2.2.3b in Concrete Bridges having Multi Lane with More than 4 Main Girders & Linked by

Factor as mentioned in Equation of Table- 4.6.2.2.3a-2.

Girder of Multi Lane RCC T-Girder Bridge is being Expressed by gExt-V = egInt-V ; where,

e is Correction Factor having value = 0.60 + de/3000; Here ,

de is a Distance in mm from Exterior Face/Web of Exterior Girder up to the de

Interior Edge of Curb or Traffic Barrier = WEdge-to-Curb. - SCant.

Since Position of Exterior Face/Web of Exterior Girder is within Inboard of

Slab's Outer Edge-to-Curb, thus de is of (+) ve value.

Thus Calculated value of Distribution Factor for Overhanging gExt.= egInt-V gExt.-V

(Having value of de = 575 > (-)300 but < 1700; ts = 200 > 110 but < 300 ;


a)the Live Load (Truck/Tandem) Moments for Interior Deck Spans should be Calculated in respect of the Strip Width

b)

for (-) ve Moment is being Expressed by the Equation, (-)XInt. = 1220 + 0.25S in mm. Here,

c) S 2,000.000 mm

d) Thus Equivalent Strip Width for (+) ve Moment 1,760.000 mm

e) Thus Equivalent Strip Width for (-) ve Moment 1,720.000 mm

vii) Computation of Equivalent Strip Width for Live Load Moment for Deck Overhanging under Provisions ofAASHTO-LRFD-4.6.2.1.3 :

a)Load & Calculation of respective Moments on Cantilever/Overhanging Portion of Deck Slab, thus Provisions of thisArticle is not Applicable.

12 Moment Arms for Calculation of Moments at Different Sections due to Applied Loads :

i) Computation of Moment Arms for the Components Resting upon the Overhanging/Cantilever Deck Slab & Live Loads in Respect of Section 1-1 (On outer Face of Exterior Girder) :

a) 0.862 m

b) 0.823 m

c) 0.813 m

d) 0.100 m

e) 0.212 m

f) 0.375 m

g) 0.700 m

ii) Computation of Moment Arms for the Components Resting upon the Deck Slab on Inner Side of Exterior Girder & Live Loads in Respect of Section 2-2 (On Inner Face of Exterior Girder) :

a) 0.088 m

b) 0.088 m

According to Article - 4.6.2.1.3 in a Concrete Bridges having More than 4 Main Girders & Linked by Cross-Girders,

as mentioned in Equation of Table- 4.6.2.1.3-1.

For Cast in Place PC-Girder Concrete Bridge with Considered Strip Perpendicular to Traffic, Equivalent Strip Width

for (+) ve Moment is being Expressed by the Equation, (+)XInt. = 660 + 0.55S in mm.& that

S is Spacing of Supporting Components in mm (Here PC-Girders) = LC/C-Gir.

(+)XInt.

(-)XInt.

Since the Provisions of Article-4.6.2.1.3 have Recommended to use Provisions of Article-3.6.1.3.4 for Applied Live

Moment Arm for Railing = LCant-Ext. -bRailing/2 LRailing

Moment Arm for Railing Post = LCant-Ext. - WPost-Trans./2 LR-Post

Moment Arm for Raining Post Base = LCant-Ext. - bCurb/Guard./2 LR-Post-Base

Moment Arm for Sidewalk Slab = WSidewalk/2 - bGir-Main - LCurb-inner-Ext. LS-walk

Moment Arm for Utility/Fluid Space = WFluid/2 - LCurb-Outer-Ext. LUtility.

Moment Arm for Pedestrian Live Load = WSidewalk/2 - LCurb-Outer-Ext. LPedes

Moment Arm for Line Live Load = WSidewalk - LCurb-Outer-Ext. - 0.300 LLine

Moment Arm for Curb/Wheel Guard (Part) = LCurb-inner-Ext./2 LCurb

Moment Arm for Pedestrian Live Load on Curb = LCurb-Inner-Ext. LPedes-Curb

c) Moment Arm for Wearing Course & Lane Live Load on Deck in between 0.913 m

13 Calculation of (-) Moments (DL & LL) of Overhanging Span at Section 1-1 (Exterior Girder Outer Face) Based Distribution Factor or Lever Rule under Provisions Strength Limit State (USD); AASHTO-LRFD-2004 :

i) Since the Span is Overhanging/Cantilever, thus Moment value is of (-) ve in Nature.

ii) Calculation of Factored Dead Load (DL) Moments for Overhanging/Cantilever Span at Section 1-1 (Outer Face of Exterior Girder) :

a) 2.708 kN-m/m

b) 1.902 kN-m/m

c) 0.668 kN-m/m

d) 2.011 kN-m/m

e) Moment due to Curb/Wheel Guard, since it is Outboard of Overhanging thus calculation of Moment is not required.

f) 0.225 kN-m/m

g) 0.663 kN-m/m

h) 8.177 kN-m/mOverhanging.

iii) Calculation of Factored Live Load (LL) Moments for Overhanging/Cantilever Span at Section 1-1 (Outer Face of Exterior Girder) :

a) 5.512 kN-m/m

b) 17.885 kN-m/m

c) 23.397 kN-m/mOverhanging.

iv) 31.574 kN-m/mApplied Loads (DL + LL) on Overhanging/Cantilever Span of Deck.

14Distribution Factor or Lever Rule under Provisions Strength Limit State (USD); AASHTO-LRFD-2004 :

i) For Continuous Span at Support Position Moment value is of (-) ve in Nature.

LWC/Lane

Exterior Girder & 1st. Interior Girder = (SInter - LCurb-Inner-Ext.)/2 + LCurb-Inner-Ext.

Moment due to Self Weight of Deck Slab = WDL-Self.-USD*(SCanti.)2/2 MDL-Self-USD

Moment due to Railing = WDL-Railing-USD*LRailing MDL-Railing-USD

Moment due to Railing Post = WDL-R-Post-USD*LR-Post MDL-R-Post-USD

Moment due to Railing Post Base = WDL-R-Post-Base-USD*LR-Post-Base MDL-R-Post-base-USD

Moment due to Sidewalk Slab = WDL-S-Walk.-USD*LS-Walk. MDL-S-Walk.-USD

Moment due to Utility = WDL-Utility.-USD*LUtility. MDL-Utility.-USD

Total Factored Dead Load (-) Moments at Section 1-1 for (-)MDL-Total-1-1-USD

Moment due to Pedestrian on Sidewalk = PLL-Pedst.-USD*WSidewalk*LPedes MLL-Pedst-USD

Moment due to Line load on Sidewalk = PLL-Line.-USD*LLine MLL-Line-USD

Total Factored Live Load (-) Moments at Section 1-1 for (-) MLL-Total-1-1-USD

Calculated Total Factored (-)Moments at Section 1-1 due to (-) åMDL+LL-1-1-USD

Calculation of (-) Moments (DL & LL) of Interior Span at Section 2-2 (Exterior Girder Inner Face) Based

ii) Calculation of Factored Dead Load (DL) Moments Interior Continuous Span at Section 2-2 (Inner Face of Exterior Girder) :

a) 1.815 kN-m/m

b) 0.783 kN-m/m

c) 0.217 kN-m/m

d) 2.814 kN-m/mSpan.

iii) Calculation of Factored Live Load (DL) Moments Interior Continuous Span at Section 2-2 (Inner Face of Exterior Girder) :

a) 82.849 kN-m/m

b) 0.938 kN-m/m

c) 83.787 kN-m/m

iv) 86.602 kN-m/m Applied Loads (DL + LL) on Interior Span of Deck.

15Based Distribution Factor or Lever Rule under Provisions Strength Limit State (USD); AASHTO-LRFD-2004 :

i) For Continuous Span at Middle Position the Moment value is of (+) ve in Nature.

ii) Calculation of Factored Dead Load (DL) Moments Interior Continuous Span at Section 3-3 (On Middle of Interior Girders) :

a) 1.167 kN-m/m

b) 0.503 kN-m/m

c) 1.670 kN-m/mSpan.

iii) Calculation of Factored Live Load (DL) Moments Interior Continuous Span at Section 3-3 (On Middle of Interior Girders) :

a) 82.849 kN-m/m

b) 0.603 kN-m/m

c) 83.452 kN-m/m

Moment due to Self Weight of Deck Slab = WDL-Self.-USD*(SInter.)2/9 MDL-Self-USD

Moment due to Wearing Course = WDL-WC.-USD*(SInter)2/9 MDL-WC-USD

Moment due to Curb/Wheel Guard Portion , = WDL-Curb.-USD*LCurb. MDL-Curb.-USD

Total Factored Dead Load (-) Moments at Section 2-2 for Interior (-)MDL-Total-2-2-USD

Moment due to Live Truck Wheel Load = PLL-Truck.-USD*gInt.-M MLL-Truck-USD

Moment due to Live Lane load on Sidewalk = PLL-Lane.-USD*(SInter.)2/9 MLL-Lane-USD

Total Factored Live Load (-)Moments at Section 2-2 for Interior Span (-)MLL-Total-2-2-USD

Calculated Total Factored (-) Moments at Section 2-2 due to (-)åMDL+LL-2-2-USD

Calculation of (+) Moments (DL & LL) of Interior Span at Section 3-3 (On Middle between Interior Girders)

Moment due to Self Weight of Deck Slab = WDL-Self.-USD*(SInter.)2/14 MDL-Self-USD


Total Factored Dead Load (+) Moments at Section 3-3 for Interior (+)MDL-Total-3-3-USD


Moment due to Live Lane load on Sidewalk = FPLL-Lane.-USD*(SInter.)2/9 MLL-Lane-USD

Total Factored Live Load (+)Moments at Section 3-3 for Interior (+)MLL-Total-3-3-USD

Span.

iv) 85.122 kN-m/m Applied Loads (DL + LL) on Interior Span of Deck.

16Distribution Factor or Lever Rule under Provisions Strength Limit State (USD); AASHTO-LRFD-2004 :


ii) Calculation of Factored Dead Load (DL) Moments Interior Continuous Span at Section 4-4(On Face of Interior Girder) :

a) 1.815 kN-m/m

b) 0.783 kN-m/m



a) 82.849 kN-m/m

b) 0.938 kN-m/m


iv) 86.385 kN-m/m

17

i) Section on Which Calculated (-) Moment is Highest; Section 2-2 is highest

ii) The Calculated value of Highest (-) ve Moment 86.602 kN-m/m

18Bridge Deck Slab from the Calculated Values of Approximate Method & Distribution Factor Method under Provisions of Strength Limit State (USD); AASHTO-LRFD-2004 :

i) Distribution Factor Method Highest

ii) Approximate Method Higer

Calculated Total Factored (+) Moments at Section 3-3 due to (+)åMDL+LL-3-3-USD

Calculation of (-) Moments (DL & LL) of Interior Span at Section 4-4 (On Face of Interior Girder) Based

Moment due to Self Weight of Deck Slab = WDL-Self-USD.*(SInter.)2/9 MDL-Self-USD


Total Factored Dead Load (-) Moments at Section 4-4 for Interior (-)MDL-Total-4-4-USD


Moment due to Live Lane load on Sidewalk = PLL-Lane.-USD*(SInter.)2/9 MLL-Lane-USD

Total Factored Live Load (-)Moments at Section 4-4 for Interior (-)MLL-Total-4-4-USD

Calculated Total Factored (-) Moments at Section 4-4 due to (-)åMDL+LL-4-4-USD

Applied Loads (DL + LL) on Interior Span of Deck.

Calculated Highest (-) ve Moment value for Distribution Factor Method under Strength Limit Stalt (USD):

(-)åMDL+LL-2-2-USD

Selecton of Approprate (-) Moments & (+) Moments (DL+LL) for Flexural Design of Reinforcements of the

The Highest of Calculated (+) ve Moment Either Approximate Method or Distribution Factor Method.

The Highest of Calculated (-) ve Moment Either Approximate Method or Distribution Factor Method.

iii)at Sections 3-3 & that for (-) ve Moments at Section 2-2 both are Heigher than the Calculated Factored Moments According to Provisions under Approximate Methods. Thus the Calculated Factored Moments According to Provisions of Distribution Factor Methods at Section 3-3 & 2-2 are the Governing Moments for Flexural Design of Deck Slab Reinforcements.

iv) 85.122 kN-m/m

v) 87.014 kN-m/m

18Based Distribution Factor or Lever Rule under Provisions Service Limit State (WSD); AASHTO-LRFD-2004 :

i)

ii) Calculation of Factored Dead Load (DL) Moments Interior Continuous Span at Section 3-3 (On Middle of Interior Girders) :

a) 0.933 kN-m/m

b) 0.335 kN-m/m


iii) Calculation of Factored Live Load (DL) Moments Interior Continuous Span at Section 3-3 (On Middle of Interior Girders) :

a) 35.596 kN-m/m

b) 0.603 kN-m/m


iv) 37.468 kN-m/m

19Distribution Factor or Lever Rule under Provisions Service Limit State (WSD); AASHTO-LRFD-2004 :


ii) Calculation of Factored Dead Load (DL) Moments Interior Continuous Span at Section 4-4(On Face of Interior Girder) :

The Calculated Values of Factored Moments according to Distribution Factor Methods for (+) ve Moments

Total Factored Flexural Design (+) ve Moments (DL + LL) for Deck (+)MDesign-USD

Total Factored Flexural Design (-) ve Moments (DL + LL) for Deck (-)MDesign-USD

Calculation of (+) Moments (DL & LL) of Interior Span at Section 3-3 (On Middle between Interior Girders)

For Continuous Span at Middle Position the Moment value is of (+) ve in Nature.

Moment due to Self Weight of Deck Slab = WDL-Self.-WSD*(SInter.)2/14 MDL-Self-WSD

Moment due to Wearing Course = WDL-WC.-WSD*(SInter)2/14 MDL-WC-WSD

Total Factored Dead Load (+) Moments at Section 3-3 of Interior (+)MDL-Total-3-3-WSD

Moment due to Live Truck Wheel Load = PLL-Truck.-WSD*gInt.-M MLL-Truck-WSD

Moment due to Live Lane load on Sidewalk = PLL-Lane.-WSD*(SInter.)2/9 MLL-Lane-WSD

Total Factored Live Load (+)Moments at Section 3-3 for Interior (+)MLL-Total-3-3-WSD

Calculated Total Factored (+) Moments at Section 3-3 due (+)åMSDL+LL-3-3-WSD

to Applied Loads (DL + LL) on Interior Span of Deck.

Calculation of (-) Moments (DL & LL) of Interior Span at Section 4-4 (On Face of Interior Girder) Based

a) 1.452 kN-m/m

b) 0.522 kN-m/m

c) 1.974 kN-m/mSpan


a) 35.596 kN-m/m

b) 0.938 kN-m/m


iv) 38.507 kN-m/m

20

i) For Continuous Span at Middle Position the Moment value is of (+) ve & on Support is (-) ve.

ii) Calculation of Unfactored Dead Load (+)Moments of Interior Continuous Span at Middle :

a) 0.933 kN-m/m

b) 0.335 kN-m/m

c) Total Factored Dead Load (+) Moments at Middle of Interior Span. 1.269 kN-m/m

iii) Calculation of Unfactored Dead Load (-)Moments of Interior Continuous Span at Support :

a) 1.452 kN-m/m

b) 0.522 kN-m/m

c) Total Unfactored Dead Load (-) Moments at Middle of Interior Span. 1.974 kN-m/m

21 Calculation of Factored Shearing Forces on Interior Span Faces According to Distribution Factor & Lever Rule under Provisions Strength Limit State (USD);

i) Calculation of Factored Dead Load (DL) Shearing Forces on Faces of Interior Span at Grider Faces :

a) 4.950 kN/m

Moment due to Self Weight of Deck Slab = WDL-Self.-WSD*(SInter.)2/9 MDL-Self-WSD

Moment due to Wearing Course = WDL-WC.*(SInter)2/9 MDL-WC-WSD

Total Factored Dead Load (-) Moments at Section 4-4 for Interior (-)MDL-Total-2-2-WSD

Moment due to Live Truck Wheel Load = PLL-Truck.-WSD*gInt.-M MLL-Truck-WSD

Moment due to Live Lane load on Sidewalk = PLL-Lane.-WSD*(SInter.)2/9 MLL-Lane-WSD

Total Factored Live Load (-)Moments at Section 4-4 for Interior (-)MLL-Total-4-4-WSD

Calculated Total Factored (-) Moments at Section 4-4 due to (-)åMSDL+LL-4-4-WSD

Applied Loads (DL + LL) on Interior Span of Deck.

Unfactored (+) ve & (-) ve Dead Load Moments under Provisions of AASHTO-LRFD-2004 :

Moment due to Self Weight of Deck Slab = WDL-Self.*(SInter.)2/14 MDL-Self-UF

Moment due to Wearing Course = WDL-WC.*(SInter)2/14 MDL-WC-UF

(+)MDL-Total-UF

Moment due to Self Weight of Deck Slab = WDL-Self.*(SInter.)2/9 MDL-Self-UF

Moment due to Wearing Course = WDL-WC.*(SInter)2/9 MDL-WC-UF

(-)MDL-Total-UF

Shearing Forces due to Self Weight of Deck Slab = WDL-Self.-USD*SInter./2 VDL-Self--USD

b) 2.135 kN-m/m

c) Total Factored Dead Load (-) Shearin Forcets at Faces of Interior Girder. 7.085 kN-m/m

ii) Calculation of Factored Live Load (DL) Shearing Forces on Faces of Interior Span at Grider Faces :

a) 107.077 kN/m

b) 4.476 kN/m

c) 111.553 kN/m

iii) Tatal Factored Shearing Forces On Interior Girder Feces due to 118.637 kN/mApplied Loads :

22 Computation of Related Features required for Flexural Design of Top & Bottom Reinforcements for Deck Slab Against Calculated Design (-) ve & (+) ve Moments :

i) Design Strip Width for Deck Slab in Transverse Horizontal Direction & Clear Cover on Different Faces:

a) Let Consider the Design Width in Transverse Directions is = 1000mm b 1.000 m

b) 25 mm

38 mm

ii) Calculations of Limits For Maximum Reinforcement, (AASHTO-LRFD-5.7.3.3.1) :.

a) With Maximum Amount of Prestressed & Nonprestressed Reinforcement for 0.42

b) c Variable

c) Variable

Variable

Variable

410.00

Variable

Variable mm

Variable mm

d) For a Structure having only Nonprestressed Tensial Reinforcement the values of

Shearing Forces due to Wearing Course = WDL-WC.-USD*SInter/2 VDL-WC-USD

VDL-Total-USD

Shearing Forces due to Live Truck Wheel Load = PLL-Truck-USD.*gInt.-V VLL-Truck-USD

Shearing Forces due to Live Lane Load = FSPLL-Lane.*SInter./2 VLL-Lane-USD

Total Factored Live Load (-)Moments at Section 4-4 for Interior Span VLL-Total-USD

åVDL+LL-USD

Let the Clear Cover on Bottom Surface of Decl Slab, C-Cov.Bot. = 25mm, C-Cov-Bot.

Let the Clear Cover on Top Surface of Decl Slab, C-Cov.Top = 38mm, C-Cov-Top.

c/de-Max.

a Section c/de £ 0.42 in which;

c is the distance from extreme Compression Fiber to the Neutral Axis in mm

de is the corresponding Effective Depth from extreme Compression Fiber to de

the Centroid of Tensial Forces in Tensial Reinforcements in mm. Here;







Centroid in mm.

xii) ds = Distance of Centroid of Nonprestressed Tensial Reinforcement from ds

the Extreme Compression Fiber in mm.


iii) Limits For Manimum Reinforcement, (AASHTO-LRFD-5.7.3.3.2) :

a) For Section of a Flexural Component having both Prestressed & Nonprestressed Tensile Reinforcements should

b) Variable N-mmwhere;

- Extreme Fiber only where Tensile Stress is caused by Externally Applied

Variable N-mm

Variable

0.006667

6.667/10^3

6.667*10^6

2.887

c) 19.247 kN-m

19246817.919 N-mm

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)

Position Value of Value of Actuat Acceptable M Maximum

& Nature Unfactored Cracking Factored Allowable Flexuralof Moment Dead Load As per Moment Cracking Cracking Moment Factored Min. Moment Moment

on Moment Equation Value Moment Moment of Section Moment

Back 5.7.3.3.2-1 M (1.33*M)Wall kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

thus de = ds .

have Minimum Resisting Moment Mr ³ 1.2*Mcr or 1.33 Times the Calculated Factored Moment for the Section Based on AASHTO-LRFD-3.4.1-Table-3.4.1-1, which one is less.For Compnents having Nonprestressed Tensile

Reinforcements only Mr = 1.2Mcr.

The Cracking Moment of a Section Mcr = Sc(fr + fcpe) - Mdnc(Sc/Snc -1) £ Scfr Mcr

i) fcpe = Compressive Stress in Concrete due to Effective Prestress Forces at fcpe N/mm2

Forces after allowance of all Prestressing Losses in MPa. In Nonprestressing

RCC Components value of fcpe = 0.



iii) Sc = Section Modulus for the Extreme Fiber of the Composite Section Sc mm3

where Tensile Stress Caused by Externally Applied Loads in mm3.

iv) Snc = Section Modulus of Extreme Fiber of the Monolithic/Noncomposite Snc m3

Section where Tensile Stress Caused by Externally Applied Loads in mm3. m3

For the Rectangular RCC Section value of mm3

Snc = (b*tSlab.3/12)/(tSlab./2)

v) fr = Modulus of Rupture of Concrete in Mpa,(AASHTO LRFD-5.4.2.6). fr N/mm2





Cpoputed value of Mcr = 1.33*MExt Factored Moment due to External Forces Mcr-3

Table-1 Showing Allowable Resistance Moment M r for Minimum Reinforcement of Different Surface & Direction

1.2 Times 1.33 Times Mr

Mcr-1 Mcr of Mcr of M,

for RCC Mu

MDL-UF Sncfr (Mcr-1£Sncfr) (1.2*Mcr) 1.2Mcr (M ³ Mr)

1.269 19.247 19.247 19.247 23.096 87.014 115.728 23.096 87.014

of Girder

(+)ve Mid. 1.974 19.247 19.247 19.247 23.096 85.122 113.213 23.096 85.122

of Span

iv)

a) Balanced Steel Ratio or the Section, 0.022

b) 0.016

23on Interior Span Strip :

i) Design Moment for the Section :

a) The Calculated Flexural (-) ve Moment on Interior Span Strip of 87.014 kN-m/m

87.014*10^6 N-mm/m

Moment Governs the Provision of Reinforcement against (-) ve Moment 23.096 kN-m/mvalue. For (-) ve value the required Reinforcement will be on Top Surface. 23.096*10^6 N-mm/m

b) 87.014 kN-m/m

87.014*10^6 N-mm/m

ii) Provision of Reinforcement for the Section :

a) 16 mm

b) 201.062

c) The provided Effective Depth for the Section with Reinforcement on Top 154.000 mm

d) 35.820 mm

e) 1,732.750

f) 116.036 mm,C/C

g) 100 mm,C/C

h) 2,010.619

(-)ve Face

Calculations for Balanced Steel Ratio- pb & Max. Steel Ratio- pmax according to AASHTO-1996-8.16.2.2 :

pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),

Max. Steel Ratio, pmax. = f *pb , (Here f = 0.75) pmax.

Flexural Design of Reinforcements on Bottom Surface of Deck Slab Against Calculated (-) ve Moment

(-)åMDL+LL-USD

Deck Slab is Greater than the Allowable Minimum Moment Mr. Thus Calculated

Mr

Since (-)åMDL+LL-USD > Mr, the Allowable Minimum Moment for the MU

Section thus MDL+LL-USD is the Design Moment MU.

Let provide 16f Bars as Reinforcement on Top Surface of Deck Slab DDect-Top

X-Sectional of 16f Bars = p*DDect-Top2/4 Af-16. mm2

de-pro.

Surface, dpro = (tSlab-CCov-Top. -DDect-Top/2)

With Design Moment MU , Design Strip Width b & Effective Depth dpro; areq.

the required value of a = dpro*(1 - (1 - (2MU)/(b1f/cbdpro

2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Top mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Top sreq

Let the provided Spacing of Reinforcement with 16f bars for the Section spro.

spro = 100mm,C/C

The provided Steel Area with 16f bars having Spacing 100mm,C/C As-pro-Top. mm2/m

= Af-16.b/spro

iii) Chacking in respect of Design Moment & Max. Steel Ratio :

a) 0.013

b) 46.182 mm

c) Resisting Moment for the Section with provided Steel Area, 107.915 kN-m/m

d) Mpro>Mu OK

e) p-pro<p-max OK

iv) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

a) 0.450

b) c 39.255 mm

c) 0.850

d) 0.255

n) c/de-pro<c/de-max. OK

v) Checking Against Max. Shear Force on Deck Slab.

a) The Maximum Shear Force occurs at Face of Longitudinal Girder on Deck 118.637 kN/mSlab Span Strip which is also the Ultimate Shearing Force for the Section. 118.637*10^3 N/m

b)

b-i) 1,000.000 mm

a-ii) 144.000 the neutral axis between Resultants of the Tensile & Compressive Forces due

138.600 mm 144.000 mm

b-iii) f 0.900

Steel Ratio for the Section, ppro = As-pro/bdpro ppro

With provided Steel Area the value of 'a' = As-pro*fy/(b1*f/c*b) apro

Mpro

= As-pro*fy(d - apro/2)/10^6

Relation between Resisting Moment Mpro & Designed Moment MU.

Relation between Provided Steel Ration rpro & Allowable Max. Steel Ratio rMax.

Accodring to AASHTO-LRFD-.7.3.3.1; In Flexural Design c/de £ 0.42; where, c/de-Max.

c is the Distance between Neutral Axis& the Extrime Compressive Face,

having c = b1apro, in mm.

b1 is Factor for Rectangular Stress Block for Flexural Design b1

Thus for the Section the Ratio c/de = 0.276 c/de-pro

Relation between c/de-Max. & c/de-pro (Whether c/de-pro< c/de-Max. or Not)

VU.

Thus Maximum Shear Force, VMax = SVDL+LL-USD = VU

The Shearing Stress on Concrete due to Applied Shear Force at a Section. vu = (VU - fVp)/fbvdv, (AASSHTO-LRFD-5.8.2.9).Here,

bv is Minimum Width of the Section, here bv = b, the Design Strip Width. bv.

dv is Effective Shear Depth taken as the distance measured perpendicular to dv.

to Flexural having value = 0.9de or 0.72h in mm, which one is greater.

Where; de = dpro the provided Effective Depth of Tensile Reinforcement &

h = tSlab Thickness of Deck Slab.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h

f is Resistance Factor for Shear

b-iv) - N

c) 756.000 kN/m 756.000*10^3 N/m

1,095.419 kN/m 1095.419*10^3 N/m

756.000 kN/m 756.000*10^3 N/m

c-i) 1,095.419 kN/m(AASHTO-LRFD- Equ. 5.8.3.3-1); 1095.419*10^3 N/m

c-ii) - N/m

c-iii) b 2.000

c-iv) - N

d) Vn>Vu Satisfied

e)thus the Abutment Wall does not require any Shear Reinforcement.

f)Bottom Section does not Require any Shear Reinforcement, thus Flexural Design of Vertical Reinforcement on

vi) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 97.124 N-mmwhere; 97.124*10^6 kN-m

107.915 N-mm 107.915*10^6 kN-m

f 0.90

b)

Vp is component of Prestressing Force in direction of Shear Force in N; Vp.

(Sinec the Well Cap is a RCC Structure, thus Vp = 0.

The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-Deck of Equations as mentioned in Aritical 5.8.3.3 :

i) Vn-1 = Vc + Vs + Vp Equ.- 5.8.3.3-1, or Vn-1

ii) Vn-2 = 0.25f/cbvdv + Vp Equ.- 5.8.3.3-2. In which, Vn-2

Vc is Nominal Shear Resistance of Conrete in N & value = 0.083bÖf/cbvdv, Vc

Vs is Shear Resistance Provided by Shear Reinforcement in N having value Vs

= Avfydv(cotq + cota)sina /s. (AASHTO-LRFD-Equ. 5.8.3.3-3) in which,

For Footing/Foundation/Slab Vs = 0.

b is Factor for the Diagonally Cracked Concrete to transmit Tension as per AASHTO-LRFD-5.8.3.4. For Footing/Foundation/Slab b = 2.00.


(For RCC Structure Elements, Vp = 0. AASHTO-8.16.6.3.1.)

Statue between Computed Nominal Shear Resitance Vn & Factored Shearing Forces

VU for the Section (Whether Vn > VU or Vn < VU & Provisions of AASHTO-LRFD-5.8.3 have Satisfied or Not).

Since Nominal Shear Resitance for the Section Vn > VU the Calculated Ultimate Shearing Force for the Section,

Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Back Wall on its

Earth Face of Back Wall is OK.

Factored Flexural Resistance for any Section of Component, Mr = fMn, Mr


ii) f is Resistance Factor of Flexural in Tension of Reinforcement/Prestressing.

The Nominal Resistance of Rectangular Section with One Axis Stress having both Prestressing & Nonprestessing

c) In a Nonprestressing Structural Component having Rectangular Elements, at any Section the Nominal Resistance,

d) Since Deck Slab is being a Continuous Sturucture & for Design purpose it 107.915 kN-mis being considered as Constituents of 1.000 m Wide Strips. The Steel Area 107.915*10^6 N-mmagainst Factored Max. Moments on Main Girder Face will have the value of

e) 87.014 kN-m

87.014*10^6 N-mm

f) Mr>Mu Satisfied

vii) Checking in respect of Control of Cracking By Distribution of Reinforcement, (AASHTO-LRFD-5.7.3.4) :

a)

Where;

b) 124.364

38.507 kN-m at Section 4-4, which is the Highest one of (-) WSD values. 38.507*10^6 N-mm

2,010.619

154.000 mm the Tensile Reinforcement for the Section.

c) 1.419

58.000 mmTension Bar. The Depth is Summation Earth/Water Clear Cover & Radius of the

A 11,600.000 by Dividing the Total Concrete Area bounded in between Extreme Tension Face & a Straight Line parallel to Neutral Axis of Component having equal distance from

AASHTO-LRFD-5.7.3.2.3 is Mn = Apsfps(dp-a/2) + Asfy(ds-a/2) - A/sf/

y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-Top

Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment MU on Face of Main Girder of SMDL+LL-USD

Continuous Span Strip = (-) åMDL+LL-USD

Relation between the Computed Factored Flexural Resistance Mr & the Actual

Factored Moment MU at Mid Span ( Which one is Greater, if Mr ³ MU the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

Under Service Limit State Load Condition, Developed Tensile Stress of Reinforcement fs-Dev. of Concrete Elements,

should not exceed fs the Computed Tensile Stress of Reinforcement under provision of AASHTO-LRFD-5.7.3.4.

fs-Dev. is Developed Tensile Stress in Provided Reinforcements of Section fs-Dev. N/mm2

under the Service Limit State of Loads = M/As-prode in which,

i) M is Calculated Moment for the Section under Service Limit State (-)åMDL+LL-4-4-WSD

ii) As-pro is the Steel Area for the Section under USD Design Calculation. As-pro mm2

iii) de is Effective Depth between Extreme Compression Fiber to Centroid of de

fsa is Computed Tensile Stress of Reinforcement having its value fsa N/mm2

= Z/(dcA)1/3 £ 0.6fy, in Which;

i) dc= Depth of Concrete Extreme Tension Face from the Center of the Closes dc

Closest Bar to Tension Face. The Max. Clear Cover = 50mm. In a Component

of Rectangular Section, dc = DBar/2 + CCov-Top. Since Clear Cover on Earth Face of

Back Wall, CCov-Earth = 50mm & Bar Dia, DBar = 16f ; thus dc = (16/2 + 50)mm

ii) A = Area of Concrete Surrounding a Single Tension Bar, which is Calculated mm2

the Centrioed of Main Tension Reinforcement Bars on both side & Diving the Area by the total Number of Main Bars as Tensile Reinforcement having Max. Clear

Spacing between Provided Tension Bars.

23,000.000 N/mm

Since the Structure is very close to Sea, thus it’s Components are of Severe

246.000

d)

e) 124.364 N/mm

f) fs-Dev.< fs Satisfy

g) fsa< 0.6fy Satisfy

h) Zdev.< Zmax. Satisfy

i)

Width Parameter, thus Provisions of Tensile Reinforcement in Vertical on Back Wall Earth Surface in respect of

j)

m) Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio,

the Flexural Design of Cantilever Slab section of Bridge Deck Slab is OK.

23Interior Span Strip :


a) The Calculated Flexural (+ ve Moment on Interior Span Strip of Deck 85.122 kN-m/m

85.122*10^6 N-mm/m

Moment Governs the Provision of Reinforcement against (+) ve Moment 23.096 kN-m/mvalue. For (+) ve value the required Reinforcement will be on Bottom Surface. 23.096*10^6 N-mm/m

Cover = 50mm.In Abutment Wall the Tension Bars in One Layer & as per Condition

Distance of Neutral Axis from Tension Face = dc, thus Area of Concrete that

Surrounding a Single Tension Bar can Compute by A = 2dc*spro. Here spro is

iii) Z = Crack Width Parameter for Cast In Place Components in N/mm. For ZMax.

a) Structure with Moderate Exposure Components the Max. value of Z = 30000b) Structure with Severe Exposure Components the Max. value of Z = 23000c) Structure with Buried Components the Max. value of Z = 17000

Exposure Category having Allowable Max. value of ZMax. = 23000N/mm

iv) The Computed value of 0.6*fy for the Concrete Element. 0.6*fy N/mm2

Since the Calculated value of fs-Dev. is responsible for Controlling the formation of Cracks under Applied Loads to

the Back Wall Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.

Based on fs-Dve. the value of Crack Width Parameter ZDev. = fs-Dev.*(dcA)1/3 ZDev.

Relation between of Developed Tensile Stress fs-Dev. & Allowable Tensile Stress fs

Relation between Computed Tensile Stress fsa & Calculated value of 0.6fy

Relation between Allowable Max. value of ZMax. & Developed value ZDev.

Since Developed Tensile Stress of Tension Reinforcement of Back Wall fs-Dev.< fsa Computed Tensile Stress; the

Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allowable Max. Crack

Control of Cracking & Distribution of Reinforcement are OK.

More over though the Structure is a Nonprestressed one & value of dc have not Exceeds 900 mm, thus Component does require any Longitudinal Skein Reinforcement.

Flexural Design of Reinforcements on Bottom Surface of Deck Slab Against the (+) ve Moment on its

(+)åMDL+LL-3-3

Slab is Greater than the Allowable Minimum Moment Mr. Thus Calculated

Mr

b) 85.122 kN-m/m

85.122*10^6 N-mm/m


a) 16 mm

b) 201.062

c) The provided Effective Depth for the Section with Reinforcement on Water 167.000 mm

d) 31.532 mm

e) 1,525.344

f) 131.814 mm,C/C

g) 100 mm,C/C

h) 2,010.619


a) 0.012

b) 46.182 mm

c) Resisting Moment for the Section with provided Steel Area,

118.632 kN-m/m

d) Mpro>Mu OK

e) p-pro<p-max OK

iii) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

a) 0.450

b) c 39.255 mm

Since (+)åMDL+LL-3-3 > Mr, the Allowable Minimum Moment for the Section, thus MU

SMDL+LL-3-3 is the Design Moment MU.

Let provide 16f Bars as Bottom Reinforcement on Bridge Decl Slab. DBottom.

X-Sectional of 16f Bars = p*DBottom2/4 Af-16. mm2

de-pro.

Face, dpro = (tSlab.-CCov-Bot. -DBottom./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Bot. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Bot. sreq


spro = 100mm,C/C

The provided Steel Area with 16f bars having Spacing 100mm,C/C As-pro-Bot. mm2/m

= Af-16.b/spro



= As-pro*fy(d - apro/2)/10^6 Mpro

Relation between Resisting Moment Mpro & Designed Moment MU.

Relation between Provided Steel Ration rpro & Allowable Max. Steel Ratio rMax.




c) 0.850

d) 0.235

q) c/de-pro<c/de-max. OK

v) Checking Against Max. Shear Force on Deck Slab.

a) The Maximum Shear Force occurs at Face of Longitudinal Girder on Deck 118.637 kN/mSlab Span Strip which may Consider as Ultimate Shearing Force for the 118.637*10^3 N/m

b)

b-i) 1,000.000 mm


150.300 mm 720.000 mm

b-iii) f 0.900

b-iv) - N

c) 3,780.000 kN/m 3780.000*10^3 N/m

5,477.094 kN/m 5477.094*10^3 N/m

3,780.000 kN/m 3780.000*10^3 N/m


c-ii) - N/m

c-iii) b 2.000




VU.

this Section also.Thus Maximum Shear Force, VMax = SVDL+LL-USD = VU






h = tSlab Thickness of Deck Slab.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h



(Sinec the Deck Slab is a RCC Structure, thus Vp = 0.

The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-Deck of Equations as mentioned in Aritical 5.8.3.3 :







b is Factor for the Diagonally Cracked Concrete to transmit Tension as per

c-iv) - N

d) Vn>Vu Satisfied




a) 106.769 N-mmwhere; 106.769*10^6 kN-m

118.632 N-mm 118.632*10^6 kN-m

f 0.90

b)


d) Since Deck Slab is being a Continuous Sturucture & for Design purpose it 118.632 kN-mis being considered as Constituents of 1.000 m Wide Strips. The Steel Area 118.632*10^6 N-mmagainst Factored Max. Moments on Main Girder Face will have the value of

e) 85.122 kN-m

85.122*10^6 N-mm

f) Mr>Mu Satisfied


a)

AASHTO-LRFD-5.8.3.4. For Footing/Foundation/Slab b = 2.00.













y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-Bot


Calculated Factored Moment MU on Face of Main Girder of SMDL+LL-3-3-USD

Continuous Span Strip = (+) åMDL+LL-3-3-USD





Where;

b) 111.586

37.468 kN-m 37.468*10^6 N-mm

2,010.619


c) 1.273


A 11,600.000 by Dividing the Total Concrete Area bounded in between Extreme Tension Face & a Straight Line parallel to Neutral Axis of Component having equal distance fromthe Centrioed of Main Tension Reinforcement Bars on both side & Diving the Area by the total Number of Main Bars as Tensile Reinforcement having Max. Clear


23,000.000 N/mm


246.000

d)

e) 111.586 N/mm



i) M is Calculated Moment for the Section under Service Limit State (-)åMDL+LL-3-3-WSD





i) dc= Depth of Concrete Extreme Tension Face from the Center of the Closes dc


of Rectangular Section, dc = DBar/2 + CCov-Top. Since Clear Cover on Earth Face of
















i)


j)

24 Provision of Distribution Reinforcement for Bridge Dack Slab According to AASHTO-LRFD-9.7.3 :

i) Calculation of Distribution Reinforcements for Bridge Dack Slab:

a) The Bridge Deck Slab in which Reinforcements are being Provided in Primary Direction in both of its Top & Bottom

on Bottom Surface as Percentage of Primary Reinforcement against Positive Moment.

b)Main Reinforcemenr.

c)

d) For Present Case Effective Span Length = 1550 mm S 1,650.000 mm

e) Since the Primary Reinforcement Perpendicular to Traffic thus Calculated DR 94.534 %

f) Allowable Maximum % of Distribution Reinforcement 67.000 %

g) Actual % of Distribution Reinforcement 67.000 %

h) Provided Main Reinforcement Steel Area in Primary Direction on 2,010.619

i) Required Distribution Reinforcement Steel Area for Bottom Surface Against 1,347.115

j) 16.000 mm

k) 201.062




Since Developed Tensile Stress of Tension Reinforcement of Back Wall fs-Dev.< fsa Computed Tensile Stress; the




Surfaces, on its Secondary Depiction Reinforcements should arranged according to Provisions of Article -9.7.3..2

For Primary Reinforcement Parallel to Traffic the Distribution Reinforcement will be 1750/Ö(S) ≤ 50 Percent of the

For Primary Reinforcement Perpendicular to Traffic the Distribution Reinforcement will be 3840/Ö(S) ≤ 67 Percent of the Main Reinforcemenr. Where for both cases S is Effective Span Length in mm as per Article-9.7.2.3

Percent of Distribution Reinforcement ae per Formula - 3840/Ö(S)

DRMax.

DRActual-%

As-pro-Bot-Main. mm2/m

Bottom Surface of Deck Slab = As-pro-Bot.

As-Bot-Dist. mm2/m

Primary Reinforcement = As-pro-Bot-Main*DRActual*%

Let proved 16f Bars as Distribution Reinforcement for Bottom Surface DBot-Dist

X-Sectional of 16f Bars = p*DBot-Dist.2/4 Af-16 mm2

l) 149.254 mm,C/C

k) 150 mm-C/C

m) 1,340.413

n) Percentage of Provided Steel Area for Distribution Reinforcement Against 66.667 %

Provision Satisfied OK

o) Since the Provided Steel Area for Distribution Reinforcement is < 67% of Provided Main Reinforcement Steel Area, thus the Provision of Distribution Reinforcement for Deck Slab is OK.

26 Provision of Shrinkage & Temperature Reinforcements in Secondary Direction on Top Surface of Deck Slab :

a) On Deck Slab Flexural Reinforcements are being provided in Primary Direction both on Top & Bottom aginest theCalculated Moments, whereas in Secondary Direction on Bottom Surface Reinforcements are being provided underprovision of Distribution Reinforcement. Yet Reinforcements are required on Top Surface in Secondary Direction. On Top Surface of Deck Slab in Secondary Direction Reinforcements can Arrange under Provision of Shrinkage &

b) Since the Thickness of Deck Slab is less than 1200mm, thus to Calculate 1.000 m

the Shrinkage & Temperature Reinforcements on both Faces in Primary & 1.000 mSecondary Directions a Strip is being Considered having Length of each b 1.000 m

c) 268.293 Temperature Reinforcement for Structural Components having its Thickness

d) 1000000

e) 16 mmon Top Surface of Deck Slab.

f) 201.062

g) Spacing required for 16f Bars as Shrinkage & Temperature in Secondary 749.413 nos.

h)Reinforcements should not Spaced further Apart than 3.00 Times the Component's Thickness or 450mm.

600.000 mm

Spacing of Distribution Reinforcement with 16f Bars = Af-16b/As-Bot-Dist. sreq

Let the spacing of Distribution Reinforcement on Bottom with 16f bars, spro-Bot-Dist

Provided Steel Area for Distribution Reinforcement with 16f Bars having As-Dist-pro. mm2/m

150 mm C/C Spacing = Af-16.b/spro

DRpro.-%

Provided Main Reinforcement Steel Area = (As-Dist-pro./As-pro-Bot-Main)*100

Temperature Reinforcements as Mentioned in Article-5.10.8.

LPrim.

LSec.

Length of each Arm b = 1000mm.

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T mm2

1200mm or Less; As ³ 0.11Ag/fy in both way.(Here Thickness = 200mm).

Here Ag is Gross Area of Strip on Deck Slab Surface = LPrim.*LSec. Ag-Strip mm2

Let provide 16f bars as Shrinkage & Temperature in Secondary Direction DTop-S&T-Sec

X-Sectional Area of 16f bar = pDBar-S&T-V&H2/4 Af-16 mm2

sreq-S&T

Direction on Top Surface of Bridge Deck Slab = Af-16*b/As-req-S&T

According to AASHTO-LRFD-5.10.8.1. In a Component having Less 1200mm Thickness, Shrinkage & Temperature

i) 3.00 Times of Cantilever Wing Wall Thickness = 3.00*tSlab 3.00*tSlab

ii) Allowable Max. Spacing for Shrinkage & Temperature Reinforcements 450.000 mm

i) 200 mm

j) 1,005.310

k)1200mm Thickness, the Spacing of Shrinkage & Temperature Reinforcements Bars should not Exceed 300mm inEach Direction on all Faces and Steel Area of Shrinkage & Temperature Reinforcements need not Exceed value of

i) Allowable Max. Spacing for Shrinkage & Temperature Reinforcements 300.000 mm

1,500.000

l) Status between Provided Steel Area of Shrinkage & Temperature Reinforcement & Allowable Max, Steel Area for

Provisions of Shrinkage & Temperature Reinforcement have Satisfied, otherwise Not Satisfied)åAb > As-pro-S&T Satisfied

m)Shrinkage & Temperature on Surfaces of Cantilever Wing Wall is OK.

sAllow-S&T

Let provide 200 mm Spacing for Shrinkage & Temperature Reinforcements spro-S&T

with 16f Bars in Secondary Direction on Top Surface of Bridge Deck Slab.

The provided Steel Area with 16f Bars as Shrinkage & Temperature As-pro-S&T-V&H mm2/m

Reinforcements having Spacing 200mm,C/C = Af-16.b/spro-S&T

According to AASHTO-LRFD-5.10.8.1. For Components of Solid Structural Concrete Wall & Footing having Less

åAb = 0.0015Ag. Since the Bridge Deck Slab is Continuous Concrete Structure, thus

sAllow-S&T-2

ii) Calculated value of åAb = 0.0015Ag. åAb = åAb = mm2/m

Shrinkage & Temperature Reinforcement (Whether åAb > As-pro-S&T or not. If åAb < As-pro-S&T-V&H ; then

Since Calculated åAb > As-pro-S&T-V&H. > As-req-S&T-V&H. & spro-S&T-V&H. = sAllow-S&T-V&H-2, thus Provisions for the

) from Railing & Railing Post, Side

of Face of Interior Girder

. In a Component having Less 1200mm Thickness, Shrinkage & Temperature


F. Calculations for Load, Shear & Moments of RCC Main Girders under Strength Limit Stateof Design (USD) :

Description Notation DImentions Unit.

1 Structural Data :

i) Dimentions of Superstructure :

a) Span Length (Clear C/C distance between Bearings) 24.400 m

b) Addl.Length of Girder beyond Bearing Center Line. 0.300 m

c) Total Girder Length (a+2b) 25.000 m

d) Carriageway Width 7.300 m

e) Width of Side Walk on Each Side 1.250 m

f) Width of Curb/Wheel Guard 0.350 m

g) Width of Railing Curb/Post Guard 0.225 m

h) Total Width of Bridge Deck 10.250 m

i) Width & Depth of Railings 0.175 m

j) Width & Breath of Railing Post 0.225 m

k) Height of Railing Post 1.070 m

l) Height of Wheel Guard/Curb 0.300 m

m) Number of Railings on each Side 3.000 nos

n) C/C distance between Railing Posts 2.000 m

o) Thickness of Deck Slab 0.200 m

p) Thickness of Wearing Course 0.075 m

q) Number of Main Girders 5.000 nos

r) Number of Cross Girders 5.000 nos

s) Depth of Main Girders (Including Slab as T-Girder) 2.000 m

t) Depth of Cross Girders (Including Slab as T-Girder) 1.700 m

u) Width of Main Girders 0.350 m

v) Width of Cross Girders 0.250 m

w) C/C Distance between Main Girders & Flange Width 2.000 m

w-i) C/C Distance between Cross Girders in Longitudinal Direction . 6.100 m

w-ii) Distance of Slab Outer Edge to Exterior Girder Center 1.125 m

x) Clear Distance Between Main Interior Girders 1.650 m

y) Filets : i) Main Girder in Vertical Direction 0.150 m

ii) Main Girder in Horizontal Direction 0.150 m

iii) X-Girder in Vertical Direction 0.075 m

vi) X-Girder in Horizontal Direction 0.075 m

z) Vertical Surface Area of Superstructure's Exposed Elements 87.108

ii) Number of Traffic Lane on Bridge Deck:

SL

SAddl.

LGir.

WCarr-Way.

WS-Walk.

WCurb.

WR-Post.

WB-Deck.

RW&D.

PW&B.

hR-Post.

hCurb.

Rnos.

C/CD-R-Post.

tSlab.

tWC

NGirder.

NX-Girder.

hGir.

hX-Girder.

WGirder.

WX-Girder.

C/CD-Girder.

C/CD-X-Girder.

CD-Ext.-Girder-Edg.

ClD-Int.-Girder.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

ASup-Vert. m2


a) 2.028 nos

@ 2 nos (ASSHTO LRFD-3.6.1.1.1)

2 Design Data :




ii) Design AASHTO HS20 Truck Loading :









iii) Design AASHTO Lane Loading :





iv) Design AASHTO Pedestrian Loading :



v)

9.807


Number of Design Traffic Lane = WCarr.-way/3600 = 7300/3600 NLane.

Where WCarr.-way is Clear Carriageway Width in between Curbs in mm

DAxel.

DWheel.

LLRW-Load

LLRS-Load

LLMW-Load

LLMS-Load

LLFW-Load

LLFS-Load

LLLane

LLLane-Int.

LL-Pedest N/mm2




gc kg/m3






vi)






vii) Strength Data related to Ultimate Strength Design( USD & AASHTO-LRFD-2004) :

a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 200000.000 MPa

viii) Other Design Related Data :




3 Factors Applicable for Design of Different Structural Components :


a)

Here:

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wWater-Nor. kN/m3

wWater-Sali. kN/m3

wEatrh kN/m3


c

Concrete Allowable Strength under Service Load Condition (SLC) = 0.40f/c fc


c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).







VWL-Nor.

VWL-Spe.

VWA








1.000

1.000

1.000


a) The Bridge will have to face Cyclonic Storms with very high Intensity of Wind Load (Wind Velocity = 260km/hr), but

for normal wind load only are selected as CRITICAL conditions for bridge structure.

i) Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :








ii) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


b) 1.750





Qi = Force Effect,



those would be occasional. Thus the respective Multiplier Factors of Limit State STRENGTH I (Bridge used by Normal Vehicle without wind load) for normal operation, Limit State of STRENGTH-III (Wind Velocity exceeding90km/hr) for wind load during cyclonic storm condition and Limit State of STRENGTH-IV (Wind Velocity of 90 km/hr)



gEH


gEV


gES





d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

5 Load Calculations for Superstructural Components & Attachments (DL & LL) per meter Length of Girder:

i) Dead Loads on 1 no. Exterior Girder from Different Components & Attachments :




















a) Dead Load on. Exterior Girder due to Self Wt.& Attachments (Without 34.340 kN/m WC & Utilies) for per meter Length of Girder.

b) Dead Load on Exterior Girder due to WC. & Utilities for per meter 3.202 kN/m Length of Girder

c) Concentrated Dead Load on Exterior Girder from to 1 no. Cross Girder 8.526 kN

d) Sumation of Uniformly Distributed Dead Loads on Exterior Girder (a + b) for 37.543 kN/m per Meter Length of Girder.

ii) Dead Loads on 1 no. Interior Girder from Different Components & Attachments :

a) Dead Load on.Interior Girder due to Self Wt.& Attachments (Without 25.260 kN/m WC & Utilies) for per meter Length of Girder.

b) Dead Load on Interior Girder from WC. for per meter Length of Girder 3.450 kN/m

c) Concentrated Dead Load on Interior Girder from to 1 no. Cross Girder 17.053 kN

d) Sumation of Uniformly Distributed Dead Loads on Interior Girder (a + b) for 28.710 kN/m per Meter Length of Girder.

iii) Live Loads (LL) on 1 no. Exterior Girder due to Wheel Load, Lane Load & Pedestrian Load accordingProvisions of AASHTO-LRFD-3.6.1.2.2, 3.6.1.2.4 & 3.6.1.6 :

a) Sketch Diagram For Distribution of Wheel Load, Lane Load & Pedestrian Load on Exterior Girders :

Midd. & Rear Wheel Load = 72.500 kNFront Wheel Load = 17.500 kN

1.475 0.225

7.3 1.250 0.600 1.800 0.300

1.070

0.300 0.200

0.250 0.950 1.650 1.650 1.650 1.650 0.950

1.125 2.000 2.000 2.00 2.000 1.125

10.250

DLExt-Gir-Self& Atta.

DLExt-Gir-WC+ Utility.

DLExt-Gir-X-Gir.

åDLExt.U-D

DLIntt-GirSelf & Atta.

DLInt-Gir-WC.

DLInt-Gir-X-Gir.

åDL-Int.UD

CL

9.300kN/m Lane Load on 3.000m width of Deck


b)

i) From Front Wheel (Sketch Diagram) = 10.06 kN 10.063 kN

ii) From Midd. & Rear Wheel (do) = 41.69 kN 41.688 kN

c) 2.015 kN/m Full Bridge Length with Intensity of 3.100N/m/m-Wd. on 0.650m Width from Wheel Guard Face up to Middle point between Two Girders. (From Sketch Diagram) = 2.01 kN/m

d) 4.500 kN/m

Sidewalk on each side (From Sketch Diagram) = 4.50 kN/m

iv) Live Loads (LL) on 1 no. Interior Girder due to Wheel Load & Lane Load according to Provisions of AASHTO-LRFD-3.6.1.2.2, 3.6.1.2.4 & 3.6.1.6 :

a) Sketch Diagram For Distribution of Wheel Load & Lane Load on Interior Girders :

Midd. & Rear Wheel Load = 72.500 kN 72.500 Front Wheel Load = 17.500 kN 17.500

1.475 7.300 0.225

1.250 0.200 1.800 1.200 0.800 0.300

1.070

0.300 0.200

0.25 0.950 1.650 1.650 1.650 1.650 0.950

1.125 2.000 2.000 2.000 2.000 1.125

10.250

b) 26.250 kN

upon Girder & the other Line of Wheels at Axle Distance - 1.800m 108.750 kN i) Load from Front Wheel (From Sketch Diagram) = 26.250 kNii) Load from Midd. & Rear Wheel (From Sketch Diagram) = 108.750 kN

c) 6.200 kN/m Full Bridge Length having Intensity of 9.300kN/m on 3.000m Width of Deckhaving Equally distance (1.500m) from Middle point of a Girder & action for Girder with 2.000m Width (From Sketch Diagram) = 6.200 kN/m

LL on 1 no. Exterior Girder due to Wheel Load at distance 0.600m from Wheel Guard Face;

LLExt-Wheel-Front.

LLExt-Wheel-Mid& Rear.

LL on 1 no. Exterior Girder due to Lane Load Uniformly Distributed over LLExt-Lane.

LL on 1 no. Exterior Girder due to Pedestrian Load Uniformly Distributed LLExt-Pedes.

over Sidewalk on Full Bridge Length with Intensity of 4.000kN/m2 on

LL on 1 no.Interior Girder due to Wheel Load with One Line of Wheels LLInt-Wheel-Front.

LLInt-Wheel-Mid& Rear.

LL on 1 no. Interior Girder due to Uniformly Distributed Lane Load over LLInt-Lane.

CL




5 Factored Loads of Superstructure Components & Attachments (DL & LL) :

i) Factored Dead Loads on 1 no. Exterior Girder from Different Components & Attachments :

a) Factored Dead Load on Exterior Girder due to Self Wt.& Attachments 42.925 kN/m

42.925 kN/m

b) Factored Dead Load on Exterior Girder due to WC. & Utilities for 4.804 kN/m

4.804 kN/m

c) Factored Concentrated Dead Load on Exterior Girder from to 1 no. 10.658 kN

10.658 kN

d) Sumation of Factored Uniformly Distributed Dead Loads on Exterior Girder 47.729 kN/m (a + b) for per Meter Length of Girder.

ii) Factored Dead Loads on 1 no. Interior Girder from Different Components & Attachments :

a) Factored Dead Load on Interior Girder due to Self Wt.& Attachments 31.575 kN/m

31.575 kN/m

b) Factored Dead Load on Interior Girder due to WC. & Utilities for 5.175 kN/m

5.175 kN/m

c) Factored Concentrated Dead Load on Interior Girder from to 1 no. 21.316 kN

21.316 kN

d) Sumation of Factored Uniformly Distributed Dead Loads on Interior Girder 36.750 kN/m (a + b) for per Meter Length of Girder.

iii) Factored Live Loads on Different Components for 1 no. Exterior Girder :

23.420 kN = 23.420 kN

97.028 kN = 97.028 kN

b) 3.526 kN/m

3.526 kN/m

c) 7.875 kN/m

7.875 kN/m

d) 11.401 kN/m due to Lane Load & Pedestrian Loads

FDLExt-Gir-Self & Atta.

(Without WC) for per Meter Length = gDC*DLExt-Gir-Self& Atta. =

FDLExt-Gir-WC+ Utility.

per Merter Length = gDW*DLExt.-Gir-WC+Utility =

FDLExt-Gir-X-Gir.

Cross Girder = gDC*DLExt-Gir-X-Gir. =.

åFDL-Ext.UD

FDLIntt-GirSelf & Atta.

(Without WC) for per Meter Length = gDC*DLInt-Gir-Self& Atta. =

FDLInt-Gir-WC+ Utility.

per Merter Length = gDW*DLInt.-Gir-WC =

FDLInt-Gir-X-Gir.

Cross Girder = gDC*DLInt-Gir-X-Gir. =.

åFDL-Int.UD

i) From Front Wheel = mgLL-Truck*IM*LLExt-Wheel-Front FLLExt-Wheel-Front.

ii) Load from Midd.& Rear Wheel=mgLL-Truck*IM*LLExt-Wheel-Mid&Rear. FLLExt-Wheel-Mid& Rear.

Factored LL on 1 no. Exterior Girder due to Lane Load FLLExt-Lane.

= mgLL-Lane*LLExt-Lane. =

Factored LL on 1 no. Exterior Girder due to Pedestrian Load FLLExt-Pedes.

= mgLL-PL*LLExt-Pedes. =

Summation of Factored LL of Exterior Girder for per meter Length åFLL-Ext.


iv) Factored Live Loads of Different Components for 1 no. Interior Girder :

a) 61.097 kN

= 61.097 kN

253.116 kN = 253.116 kN

b) Factored LL on 1 no. Interior Girder due to Lane Load for per meter Length 10.850 kN/m

10.850 kN/m

6 Shear & Moments at different Positions of an Exterior Girder due to Factored Loads (DL& LL) from Superstructure Components & Attachments :

i) Arrangement of Wheel Loads for Exteriod Girder & c.g Point :

a)Absulate Max. Moments:

145.000 kN 145.000 kN 35.000 kN c.g. of Wheel c.g. of Girder

2.845 m 0.728 m

Rear Middle Front

4.300 4.300

b) Calculations for Center of Gravity (cg) Position of Truck with Wheel Load in 2.845 m

c)

0.728 m

d)

97.028 kN 97.028 kN 23.420 kN 2.845 c.g.

Rear Middle Front

4.300 4.300

Factored LL on 1 no.Interior Girder due to Wheel Load FLLInt-Wheel-Front.

i) Load from Front Wheel = mgLL-Truck*IM*LLInt-Wheel-Front

ii) Load from Midd.& Rear Wheel=mgLL-Truck*IM*LLInt-Wheel-Mid&Rear. FLLInt-Wheel-Mid& Rear.

FLLInt-Lane.

= mgLL-Lane*LLInt-Lane. =

Sketch Diagram Showing Wheels Loads of Truck, c.g. of Wheels & Location of Mid-Wheel under the Provisions of

c.g.Wheel

Respect of Rear Wheel; c.g. Distance from Rear Wheel

= (Wt.-Mid*4.300+Wt.-Fornt*(2*4.300))/(2*145.000+35.000)

Calculation Mid Wheel Position in Respect of Girder c.g. under Absulate Max. Moment Provision = (Distance beteen 2-Wheel - Distance of c.g. of Wheels from Rear Wheel)/2

dMid-Wheel

Sketch Diagram of Factored Wheel Loads for Exterior Girder :


e)

X-Girder-1 X-Girder-2 X-Girder-3 X-Girder-4 X-Girder-510.658 kN 10.658 kN 10.658 kN 10.658 kN 10.658 kN

0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300 m m

CL of Bearing CL of Bearing L /2

12.200 m m 3L/8 0.728 m

L /4 9.150 2.845 4.300 0.300 L /8 6.100 m m 0.300

m 3.050 m m

A B

47.729 kN/m

0.375 11.401 kN/m 0.375

m

24.40 m

25.00 m

596.610 kN 596.610 kN

26.645 kN 26.645 kN

142.516 kN 142.516 kN

115.224 102.252 kN (Max. Reaction due to Wheel Load at c.g. Position) (Max. Reaction due to Wheel Load at c.g. Position)

ii) Calculation of Factored Shear Forces & Moments at Different Locations of Exterior Girder Against AppliedFactored Loads :

a)Loads from Sidewalk, Attachments, Utilities, WC, Slab, Self Weight of Girder & Concentrated Dead Loads of Cross

Table-6-ii-a-1. Factored Shearing Forces due to all Uniformly Distributed Dead Loads on Exterior Girder.

Location From Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Shearing Forces in kN 582.291 564.393 436.718 291.146 145.573 -34.732 0.000

Table-6-ii-a-2. Factored Moments due to all Dead Loads (DL) on Exterior Girder.

Location From Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Moments in kN-m 0.000 220.373 1597.661 2751.326 3460.993 3724.446 3726.663

b) Table showing Factored Shear Forces & Moments at Different Location of Girder due to Concentrated Dead Load

Sketch Diagram of Girder with Factored Uniformly Distributed & Concentrated DL & LL, Different Locations for Shear & Moments Including Max. Reactions at Supports due to Different DL, LL-Lane Load & LL-Wheel Load :

Rear Wheel Positions under c.g Provisions. CL of Girder

Midd. Wheel Positions under c.g Provisions.Front Wheel Positions under c.g Provisions.

å FDLExt-All =

å FLLExt-Lane-Ped =

LSpan =

LTotal =

RA-DL = RB-DL =

RA-DL-X-Gir.= RB-DL-X-Gir.=

RA-LL-L = RB-LL-L =

RA-LL-Wh.= RB-LL-Wh.=

Table showing Factored Shear Forces & Moments at Different Location of Girder due to Uniformly Distributed Dead

Girders (å FDLExt-All & FDLExt-Gir-X-Gir.):

from Cross Girders (FDLExt-Gir-X-Gir.):

å FDLExt-All = 47.729 kN/m

å FLLExt-Lane-Ped. = 11.401kN/m


Table-6-ii-b-1. Factored Shearing Forces due to Concentrated Dead Loads of X-Girder on Exterior Girder.

Location From Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Shearing Forces in kN 15.987 15.987 15.987 5.329 5.329 -5.329 -5.329

Table-6-ii-b-2. Factored Moments due to all Dead Loads (DL) on Exterior Girder.


c)Loads from Sidewalk, Attachments, Utilities, WC, Slab, Self Weight of Girder & Concentrated Dead Loads of Cross

Table-6-ii-c-1. Factored Shearing Forces due to all Applied Dead Loads on Exterior Girder.

Location From Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2

Uniformly Distributed Load in kN 582.291 564.393 436.718 291.146 145.573 -34.732 0.000

Concentrated Loads in kN 15.987 15.987 15.987 5.329 5.329 -5.329 -5.329

Total Dead Load Shear in kN 598.278 580.380 452.705 296.475 150.902 -40.061 -5.329

Table-6-ii-c-2. Factored Moments due to all Applied Dead Loads on Exterior Girder.


For Uniformly Distributed in kN-m 0.000 220.373 1597.661 2751.326 3460.993 3724.446 3726.663

For Concentrated in kN-m 0.000 5.995 48.760 97.520 113.774 126.149 130.027

Total Monents in kN 0.000 226.368 1646.421 2848.846 3574.767 3850.595 3856.690

d)


Table-6-ii-d-2. Factored Moments on Exterior Girder due to Live Lane Load & Pedestrian Load (LL) .


e) Table showing Factored Shear Forces & Moments at Different Locations of Girder due to Applied Live Wheel Loads

Table-6-ii-e-1. Factored Shearing Forces on Exterior Girder due to Live Wheel-Load (LL) .

Locatio From Support-A Support On Support 0.375m L/8 L/4 3L/8 L/2



Table showing Factored Shear Forces & Moments at Different Location of Girder due to Uniformly Distributed Live

Live Load & Pedestrian Loads (åFLLExt-Lane-Ped) :

Table-6-ii-d-1. Factored Shearing Forces on Exterior Girder due to Live Lane Load & Pedestrian Load (LL) .

(FLLExt-Wheel) :


Positon of Rear/Midd. & Direction ReactionRear Wheel at Support, AB 192.122 95.094

Rear Wheel at 0.375m, AB 188.780 91.752

Rear Wheel at L/8, A B 164.937 67.910

Rear Wheel at L/4, A B 137.753 40.725

Rear Wheel at 3L/8, A B 110.568 13.541

Midd. Wheel on c.g. Poisition A B 115.224 -78.832

Rear Wheel at L/2, A B 83.384 -13.644

Midd. Wheel on c.g. Poisition B A 102.252 -18.196

Rear Wheel at L/2, BA 134.092 -83.384

Rear Wheel at 3L/8, B A 161.276 -56.200

Rear Wheel at L/4, B A 162.641 -31.415

Rear Wheel at L/8, B A 84.899 -12.128

Rear Wheel at 0.375m, B A 95.536 -1.491

Rear Wheel at Support, B A 97.028 0.000

Table-6-ii-e-2. Factored Moments on Exterior Girder due to Live Wheel-Load (LL) .


Positon of Rear/Midd. & Direction Reaction c.g.

Rear Wheel at Support, A B 192.122 0.000

Rear Wheel at 0.375m, A B 188.780 70.792

Rear Wheel at L/8, A B 164.937 503.059

Rear Wheel at L/4, A B 137.753 840.293

Rear Wheel at 3L/8, A B 110.568 1011.702

Midd. Wheel on c.g. Poisition A B 115.224 1072.359

Rear Wheel at L/2, A B 83.384 1017.285

Midd. Wheel on c.g. Poisition B A 102.252 1072.359

Rear Wheel at L/2, B A 134.092 1017.285

Rear Wheel at 3L/8, B A 161.276 857.043

Rear Wheel at L/4, B A 162.641 574.889

Rear Wheel at L/8, B A 84.899 258.943

Rear Wheel at 0.375m, B A 95.536 35.826


f)(DL), Live Loads (LL) for Lane Load & Wheel Load & their Summation for Each Point of Application :

Table-6-ii-f-1. Sum. of Factored Max. Shear Forces Against All Applied Loads (DL & LL) on Exterior Girder.

Locations from Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Loading Type Unit kN kN kN kN kN kN kN

598.278 580.380 452.705 296.475 150.902 -40.061 -5.329

139.095 134.820 104.321 69.548 34.774 -8.297 0.000

95.094 91.752 67.910 40.725 13.541 1072.359 1017.285

Total Shears on Each Point 832.468 806.951 624.936 406.747 199.216 1024.001 1011.956

e)

Table showing Max. Shear Forces at Different Locations of Exterior Girder due to respective Factored Dead Loads

a. Dead Load (åFDLExt)

b. Lane + Ped.(LL) (åFLLExt)

a. Wheel Live Load (WLLExt)

Table showing the Max. Moments at Different Locations of Exterior Girder due to respective Factored Dead Loads


(DL), Live Loads (LL) for Lane Load & Wheel Load & their Summation for Each Point of Application :

Table-6-ii-e-1. Sum. of Factored Max. Moments Against All Applied Loads (DL & LL) on Exterior Girder.

Locations from Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Loading Type Unit kN-m kN-m kN-m kN-m kN-m kN-m kN-m

0.000 226.368 1646.421 2848.846 3574.767 3850.595 3856.690

0.000 52.642 381.643 657.225 826.747 993.917 890.210

0.000 70.792 503.059 840.293 1011.702 1072.359 1017.285

Total Moments on Each Point 0.000 106.618 2531.123 4346.364 5413.216 5916.871 5764.185

7 Factored Shear & Moments at different Positions of an Interior Girder due to Factored Loads (DL& LL) from Superstructure Components & Attachments :

i) Arrangement of Wheel Loads for Interiod Girder & c.g Point :



2.845 m 0.728 m

Rear Middle Front

4.300 4.300


c)

0.728 m

d) 0.728

253.116 kN 253.116 kN 61.097 kN 2.845 c.g.

Rear Middle Front

4.300 4.300


b. Lane+Ped.(LL) (åFL&PLExt)

c. Wheel Live Load (FWLLExt)


c.g.Wheel


= (Wt.-Mid*4.300+Wt.-Fornt*(2*4.300))/(2*145.000+35.000)


dMid-Wheel

Sketch Diagram Showing Factored Wheel Loads for Interior Girder :


e)

X-Girder-1 X-Girder-2 X-Girder-3 X-Girder-4 X-Girder-5

21.316 kN 21.316 kN 21.316 kN 21.316 kN 21.316 kN

0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300

m m

CL of Bearing CL of Bearing

L /2 c.g. of Wheels.

12.200 m

m 3L/8 0.728 m

L /4 9.150 2.845 4.300

0.300 L /8 6.100 m m 0.300

m 3.050 m m

A B

36.750 kN/m

0.375 10.850 kN/m 0.375

m

24.400 m

25.000 m

459.375 kN 459.375 kN

53.290 kN 53.290 kN

135.625 kN 135.625 kN


ii) Calculation of Factored Shear Forces & Moments at Different Locations of Interior Girder Against AppliedFactored Loads :


Table-7-ii-a-1. Factored Shearing Forces due to all Uniformly Distributed Dead Loads on Interior Girder.


Table-7-ii-a-2. Factored Moments due to all Dead Loads (DL) on Interior Girder.



Sketch Diagram of Girder with Uniformly Distributed Factored DL & LL, Different Locations for Shear & MomentsIncluding Max. Reactions at Supports due to DL, LL-Lane Load & LL-Wheel Load :

Rear Wheel Positions under c.g Provisions.

CL of Girder

Midd. Wheel Positions under c.g Provisions.

Front Wheel Positions under c.g Provisions.

å FDLExt-All =


LSpan =

LTotal =

RA-DL = RB-DL =


RA-LL-L = RB-LL-L =



Girders (å FDLInt-All & FDLInt-Gir-X-Gir.):


å FDLInt-All = 36.750kN/m

å FLLInt-Lane-Ped. = 10.850kN/m


Table-7-ii-b-1. Factored Shearing Forces due to Concentrated Dead Loads of X-Girder on Interior Girder.


Table-7-ii-b-2. Factored Moments due to Concentrated Dead Loads of X-Girder on Interior Girder.



Table-7-ii-c-1. Factored Shearing Forces due to all Applied Dead Loads on Interior Girder.


For Uniformly Distributed Loads 448.350 434.569 336.263 224.175 112.088 -26.743 0.000

For Concentrated Loads 31.974 31.974 31.974 10.658 10.658 -10.658 -10.658


Table-7-ii-c-2. Factored Moments due to all Applied Dead Loads on Interior Girder.


For Uniformly Distributed Load 0.000 169.682 1230.160 2118.454 2664.880 2867.733 2869.440

For Concentrated Load 0.000 11.990 97.520 195.041 227.548 252.299 260.054

Total Monents in kN-m 0.000 181.672 1327.681 2313.495 2892.428 3120.031 3129.494

d)


Table-7-ii-d-2. Factored Moments on Interior Girder due to Live Lane Load & Pedestrian Load (LL) .



Table-7-ii-e-1. Factored Shearing Forces on Interior Girder due to Live Wheel-Load (LL) .





Live Load & Pedestrian Loads (åFLLInt-Lane-Ped) :

Table-7-ii-d-1. Factored Shearing Forces on Interior Girder due to Live Lane Load & Pedestrian Load (LL) .

(FLLInt-Wheel) :



Rear Wheel at 0.375m, AB 492.468 239.353

Rear Wheel at L/8, A B 430.272 177.156

Rear Wheel at L/4, A B 359.356 106.240

Rear Wheel at 3L/8, A B 288.439 35.324


Rear Wheel at L/2, A B 217.523 -35.592


Rear Wheel at L/2, BA 223.247 -344.081

Rear Wheel at 3L/8, B A 420.721 -146.607

Rear Wheel at L/4, B A 424.280 -81.951

Rear Wheel at L/8, B A 221.476 -31.639

Rear Wheel at 0.375m, B A 249.226 -3.890


Table-7-ii-e-2. Factored Moments on Interior Girder due to Live Wheel-Load (LL) .




Rear Wheel at 0.375m, A B 492.468 184.676

Rear Wheel at L/8, A B 430.272 1312.328

Rear Wheel at L/4, A B 359.356 2192.069

Rear Wheel at 3L/8, A B 288.439 2639.221


Rear Wheel at L/2, A B 217.523 2653.786


Rear Wheel at L/2, B A 223.247 1109.781

Rear Wheel at 3L/8, B A 420.721 2588.380

Rear Wheel at L/4, B A 424.280 2132.499

Rear Wheel at L/8, B A 221.476 675.502

Rear Wheel at 0.375m, B A 249.226 93.460



Table-7-ii-f-1. Sum. of Factored Max. Shear Forces Against All Applied Loads (DL & LL) on Interior Girder.


480.324 466.543 368.236 234.833 122.745 -37.401 -10.658

132.370 128.301 99.278 66.185 33.093 -7.895 0.000

248.072 239.353 177.156 106.240 35.324 -47.468 -35.592

Total Shears on Each Point 860.766 834.197 644.670 407.258 191.162 -92.764 -46.250

e)

Table showing Max. Shear Forces at Different Locations of Interior Girder due to respective Factored Dead Loads




Table showing the Max. Moments at Different Locations of Interior Girder due to respective Factored Dead Loads


(DL), Live Loads (LL) for Lane Load & Wheel Load & their Summation for Each Point of Application :

Table-7-ii-e-1. Sum. of Factored Max. Moments Against All Applied Loads (DL & LL) on Interior Girder.


0.000 181.672 1327.681 2313.495 2892.428 3120.031 3129.494

0.000 50.096 363.190 625.448 786.774 945.861 847.168

0.000 184.676 1312.328 2192.069 2639.221 2797.457 2653.786


8

a) d 1.883 m

c) 3.766 m

d) 352.96 kN

e) 94.768 kN

f) Shering Force due to Live Wheel Load with Rear Wheel at a Distance from 160.517 kN

g) 608.249 kN

9

i)

a) 1.712 m

Support Face.

b) Distance of Loacation from Support Point (Bearing Point), 2.087 m

c) 361.369 kN

d) 112.983 kN

e) 199.551 kN




Factored Shear Forces at a Distance 2d from Face of Support for Interior Girder :

d is Effective Depth of Neutral Axis of Tensial Reinforcement from the Extrm

Compression Fiber of T-Girder for the Section. Here d = de = 1886.000 mm.

Distance of 2d Loacation from Support Point (Bearing Point), L2d

L2d = 2de m

Shear Force due to Dead Load, FDL2d = RDL-A - å FDLInt*L2d -FDLInt-Gir-X-Gir. FDL2d

Shear Force due to Live Lane Load, FLLL2d = RLLL-A - å FLLInt*L2d FLLLdv

FWLL2d

Support Point, FWLLInt. = RWLL-A - Wheel-LL-Rear

Total Factored Shear Force at a Distance 2*d from Support, Vu-2d

FSF2d = FDL2d + FLLL2d + FWLL2d

Factored Shear Forces & Moments at a Distance d v from Face of Support for Interior Girder :

Factored Shear Forces at a Distance dv from Face of Support for Interior Girder :

dv is Effective Shear Depth of T-Girder at a Section according to provision of dv

AASHTO-LRFD-5.8.2.9 with a value dv = 1.697.000 mm for Interior Girder from

Ldv

Ldv = (L0.375+ dv) m

Shear Force due to Dead Load, FDLdv = RDL-A - å FDLInt*Ldv -FDLInt-Gir-X-Gir. FDLdv

Shear Force due to Live Lane Load, FLLLdv = RLLL-A - å FLLInt*Ldv FLLLdv

Shering Force due to Live Wheel Load with Rear Wheel at a Distance Ldv FWLLdv


f) 673.904 kN

ii)

a) 945.329 kN-m

b) 259.398 kN-m

c) Moment Wheel Loads having Rear Wheel on Critica Section 774.875 kN-m

d) Total Factored Moment at Critical Section due to Dead & Live Lodes. 1,979.602 kN-m

10 Moments at Different Locations of Exterior Girder due to Unfactored Dead Loads :

a) Sketch Diagram showing required Locations of Exterior Girder for Moments including Unfactored Dead Load.


0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300 m m

CL of Bearing CL of Bearing L /2 c.g. of Wheels.

12.20 3L /8 0.728

L /4 9.150 2.845 4.300

0.300 L /8 6.100 0.300

3.05

A B

37.543 kN/m

0.375 0.375

24.400 m

25.000 m

469.282 kN 469.282 kN

21.316 kN 21.316 kN

b) Moment at Different Section of Exterior Girder due to Uniformly Distributed & Concentrated Dead Load

Table- 10-b-1. Moments due to Unfactored Uniformly Distribute & Concentrated Dead Loads :

Locations from Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2

from Support Point, FWLLInt. = RWLL-A - Wheel-LL-Rear

Total Factored Shear Force at a Distance dv from Support Face, Vu-dv

FSFdv = FDLdv + FLLLdv + FWLLdv

Moments at a Distance dv from Face of Support for Interior Girder due to Factored Forces :

Moment due to Dead Load, = RDL-All *Ldv - åFDLInt*Ldv2/2-FDLInt-Gir-X-Gir.*Ldv Mdv-DL

Moment due to Live Lane Load, = RLLL-A *Ldv - åFLLInt*Ldv2/2 Mdv-LLL

Mdv-LWL

Mdv-DL+LL


CL of Girder



å DLExt-UDL=

LSpan =

LTotal =

RA-DL-UDL = RB-DL-UDL =

RA-DL-X-Gir = RB-DL-X-Gir =

å DLExt-UDL = 41.863kN/m


Loading Type Unit kN-m kN-m kN-m kN-m kN-m kN-m kN-m Uniformlu Distribute Load 0.000 173.341 1256.689 2164.139 2722.349 2929.576 2931.320

Concentrated Dead Load 0.000 4.796 39.008 78.016 91.019 100.919 104.022

Total Unfactored Moment 0.000 178.137 1295.697 2242.155 2813.368 3030.496 3035.342

11 Moments at Different Locations of Interior Girder due to Unfactored Dead Loads :

a) Sketch Diagram showing required Locations of Interior Girder for Moments including Unfactored Dead Load.


0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300 m m

CL of Bearing CL of Girder CL of Bearing L /2 c.g. of Wheels.

12.200 3L /8 0.728

L /4 9.150 2.845 4.300

0.300 L /8 6.100 0.300

3.050

A B

0.375 28.710 kN/m

0.375

24.400 m

25.000 m

358.875 kN 358.875 kN

42.632 kN 42.632 kN

b) Moment at Different Section of Interior Girder due to Uniformly Distributed & Concentrated Dead Load

Table- 11-b-1. Moments due to Unfactored Uniformly Distribute & Concentrated Dead Loads :

Locations from Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Loading Type Unit kN-m kN-m kN-m kN-m kN-m kN-m kN-m Uniformlu Distribute Load 0.000 132.559 961.031 1654.988 2,081.870 2240.343 2241.677

Concentrated Dead Load 0.000 9.592 78.016 156.033 182.038 226.657 208.044

Total Unfactored Moment 0.000 142.152 1039.048 1811.021 2,263.908 2,467.000 2,449.720




å DLInt-UDL=

LSpan =

LTotal =

RA-DL-UDL = RB-DL-UDL =

RA-DL-Gir = RB-DL-X-Gir =

å DLInt-UDL = 33.030kN/m


N/mm/mm-Wd.

G. Calculations for Load, Shear & Moments of RCC Main Girders under Service Limit Stateof Design (WSD) :


1 Structural Data :




















s) Depth of Main Girders (Including Deck Slab as Part of T-Girder) 2.000 m

t) Depth of Cross Girders (Including Deck Slab) 1.900 m












ii) Number of Traffic Lane on Bridge Deck:

SL

SAddl.

LGir.

WCarr-Way.

WS-Walk.

WCurb.

WR-Post.

WB-Deck.

RW&D.

PW&B.

hR-Post.

hCurb.

Rnos.

C/CD-R-Post.

tSlab.

tWC

NGirder.

NX-Girder.

hGirder.

hX-Girder.

bGirder.

bX-Girder.

C/CD-Girder.

C/CD-X-Girder.

CD-Ext.-Girder-Edg.

ClD-Int.-Girder.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

ASup-Vert. m2

a) 2.028 nos


2 Design Data :





















vi)

9.807





DAxel.

DWheel.

LLRW-Load

LLRS-Load

LLMW-Load

LLMS-Load

LLFW-Load

LLFS-Load

LLLane

LLLane-Int.

LL-Pedest N/mm2




gc kg/m3

gWC kg/m3




vii)






viii) Strength Data related to Ultimate Strength Design( USD & AASHTO-LRFD-2004) :

a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 200000.000 MPa

ix) Other Design Related Data :






a)

Here:


gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wWater-Nor. kN/m3

wWater-Sali. kN/m3

wEatrh kN/m3


c



c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).







VWL-Nor.

VWL-Spe.

VWA







1.000

1.000

1.000


a) The Bridge will have to face Cyclonic Storms with very high Intensity of Wind Load (Wind Velocity = 260km/hr), but

Normal Vehicle with wind load having Wind Velocity of 90 km/hr) for normal operation & other respective Limit State











b) 1.000

c) IM 1.000 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1(SERVICE - I);(Applicable only for Truck Loading & Tandem Loading)




Qi = Force Effect,



those would be occasional. Thus the respective Multiplier Factors of Limit State SERVICE-I (Bridge used by

SERVICE Groups are being Considered as CRITICAL conditions for Bridge Structure.



gEH


gEV


gES




d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

5 Load Calculations for Superstructural Components & Attachments (DL & LL) per meter Length of Girder:

i) Dead Loads on 1 no. Exterior Girder from Different Components & Attachments :

a) Dead Load on. Exterior Girder due to Self Wt.& Attachments (Without 34.340 kN/m WC & Utilies) for per meter Length of Girder.

b) Dead Load on Exterior Girder due to WC. & Utilities for per meter 3.202 kN/m


















DLExt-Gir-Self& Atta.

DLExt-Gir-WC+ Utility.

Length of Girder

c) Concentrated Dead Load on Exterior Girder from to 1 no. Cross Girder 8.526 kN

d) Sumation of Uniformly Distributed Dead Loads on Exterior Girder (a + b) for 37.543 kN/m per Meter Length of Girder.


a) Dead Load on.Interior Girder due to Self Wt.& Attachments (Without 25.260 kN/m WC & Utilies) for per meter Length of Girder.

b) Dead Load on Interior Girder from WC. for per meter Length of Girder 3.450 kN/m


d) Sumation of Uniformly Distributed Dead Loads on Interior Girder (a + b) for 28.710 kN/m

iii) Live Loads (LL) on 1 no. Exterior Girder due to Wheel Load, Lane Load & Pedestrian Load accordingProvisions of AASHTO-LRFD-3.6.1.2.2, 3.6.1.2.4 & 3.6.1.6 :

a) Sketch Diagram For Distribution of Wheel Load, Lane Load & Pedestrian Load on Exterior Girders :

Midd. & Rear Wheel Load = 72.500 kNFront Wheel Load = 17.500 kN

1.475 0.225

7.300 1.250 0.600 1.800 0.300

1.070

0.300 0.200

0.250 0.950 1.650 1.650 1.650 1.650 0.950

1.125 2.000 2.000 2.000 2.000 1.125

10.25

b)

i) From Front Wheel (Sketch Diagram) = 10.063 kN 10.063 kN

ii) From Midd. & Rear Wheel (do) = 41.688 kN 41.688 kN

c) 2.015 kN/m Full Bridge Length with Intensity of 3.100N/m/m-Wd. on 0.650m Width

DLExt-Gir-X-Gir.

åDLExt.U-D

DLIntt-GirSelf & Atta.

DLInt-Gir-WC.

DLInt-Gir-X-Gir.

åDL-Int.UD

LL on 1 no. Exterior Girder due to Wheel Load at distance 0.600m from Wheel Guard Face;

LLExt-Wheel-Front.

LLExt-Wheel-Mid& Rear.

LL on 1 no. Exterior Girder due to Lane Load Uniformly Distributed over LLExt-Lane.

CL


from Wheel Guard Face up to Middle point between Two Girders. (From Sketch Diagram) = 2.015 kN/m

d) 4.500 kN/m

Sidewalk on each side (From Sketch Diagram) = 4.500 kN/m

iv) Live Loads (LL) on 1 no. Interior Girder due to Wheel Load & Lane Load according to Provisions of AASHTO-LRFD-3.6.1.2.2, 3.6.1.2.4 & 3.6.1.6 :

a) Sketch Diagram For Distribution of Wheel Load & Lane Load on Interior Girders :

Midd. & Rear Wheel Load = 72.500 kN 72.500 kNFront Wheel Load = 17.500 kN 17.500 kN

1.475 7.300 0.225

1.250 0.200 1.800 1.200 0.800 0.300

1.070

0.300 0.200

0.250 0.950 1.650 1.650 1.650 1.650 0.950

1.125 2.000 2.000 2.000 2.000 1.125

10.250

b) 26.250 kN

upon Girder & the other Line of Wheels at Axle Distance - 1.800m 108.750 kN i) Load from Front Wheel (From Sketch Diagram) = 26.250 kNii) Load from Midd. & Rear Wheel (From Sketch Diagram) = 108.750 kN

c) 6.200 kN/m Full Bridge Length having Intensity of 9.300kN/m on 3.000m Width of Deckhaving Equally distance (1.500m) from Middle point of a Girder & action for Girder with 2.000m Width (From Sketch Diagram) = 6.200 kN/m

6 Factored Loads of Superstructure Components & Attachments (DL & LL) Service Limit State Design (WSD):

i) Factored Dead Loads on 1 no. Exterior Girder from Different Components & Attachments :

a) Factored Dead Load on Exterior Girder due to Self Wt.& Attachments 34.340 kN/m

34.340 kN/m

LL on 1 no. Exterior Girder due to Pedestrian Load Uniformly Distributed LLExt-Pedes.

over Sidewalk on Full Bridge Length with Intensity of 4.000kN/m2 on

LL on 1 no.Interior Girder due to Wheel Load with One Line of Wheels LLInt-Wheel-Front.

LLInt-Wheel-Mid& Rear.

LL on 1 no. Interior Girder due to Lane Load Uniformly Distributed over LLInt-Lane.

FDLExt-Gir-Self & Atta.

(Without WC) for per Meter Length = gDC*DLExt-Gir-Self& Atta. =

CL



b) Factored Dead Load on Exterior Girder due to WC. & Utilities for 3.202 kN/m

3.202 kN/m

c) Factored Concentrated Dead Load on Exterior Girder from to 1 no. 8.526 kN

8.526 kN

d) Sumation of Factored Uniformly Distributed Dead Loads on Exterior Girder 37.543 kN/m (a + b) for per Meter Length of Girder.

ii) Factored Dead Loads on 1 no. Interior Girder from Different Components & Attachments :

a) Factored Dead Load on Interior Girder due to Self Wt.& Attachments 25.260 kN/m

25.260 kN/m

b) Factored Dead Load on Interior Girder due to WC. & Utilities for 3.450 kN/m

3.450 kN/m

c) Factored Concentrated Dead Load on Interior Girder from to 1 no. 17.053 kN

17.053 kN

d) Sumation of Factored Uniformly Distributed Dead Loads on Interior Girder 28.710 kN/m (a + b) for per Meter Length of Girder.

iii) Factored Live Loads of Different Components for 1 no. Exterior Girder :

a)

10.063 kN = 10.063 kN

41.688 kN = 41.688 kN

b) 2.015 kN/m

2.015 kN/m

c) 4.500 kN/m

4.500 kN/m

d) 6.515 kN/m due to Lane Load & Pedestrian Loads

iv) Factored Live Loads of Different Components for 1 no. Interior Girder :

a) 26.250 kN

= 26.250 kN

108.750 kN

FDLExt-Gir-WC+ Utility.

per Merter Length = gDW*DLExt.-Gir-WC+Utility =

FDLExt-Gir-X-Gir.

Cross Girder = gDC*DLExt-Gir-X-Gir. =.

åFDL-Ext.UD

FDLIntt-GirSelf & Atta.

(Without WC) for per Meter Length = gDC*DLInt-Gir-Self& Atta. =

FDLInt-Gir-WC+ Utility.

per Merter Length = gDW*DLInt.-Gir-WC =

FDLInt-Gir-X-Gir.

Cross Girder = gDC*DLInt-Gir-X-Gir. =.

åFDL-Int.UD

Factored LL on 1 no. Exterior Girder due to Wheel Load at distance 0.600m from Wheel Guard Face;

i) From Front Wheel = mgLL-Truck*IM*LLExt-Wheel-Front FLLExt-Wheel-Front.

ii) Load from Midd.& Rear Wheel=mgLL-Truck*IM*LLExt-Wheel-Mid&Rear. FLLExt-Wheel-Mid& Rear.

Factored LL on 1 no. Exterior Girder due to Lane Load FLLExt-Lane.

= mgLL-Lane*LLExt-Lane. =

Factored LL on 1 no. Exterior Girder due to Pedestrian Load FLLExt-Pedes.

= mgLL-PL*LLExt-Pedes. =

Summation of Factored LL of Exterior Girder for per meter Length åFLL-Ext.

Factored LL on 1 no.Interior Girder due to Wheel Load FLLInt-Wheel-Front.

i) Load from Front Wheel = mgLL-Truck*IM*LLInt-Wheel-Front

ii) Load from Midd.& Rear Wheel=mgLL-Truck*IM*LLInt-Wheel-Mid&Rear. FLLInt-Wheel-Mid& Rear.

= 108.750 kN

b) Factored LL on 1 no. Interior Girder due to Lane Load for per meter Length 6.200 kN/m

6.20 kN/m

6 Shear & Moments at different Positions of an Exterior Girder due to Factored Loads (DL& LL) from Superstructure Components & Attachments :



2.845 m 0.728 m

Rear Middle Front

4.300 4.300


c)

0.728 m

d)

41.688 kN 41.688 kN 10.063 kN 2.845 c.g.

Rear Middle Front

4.300 4.300

e)


0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300 m m


12.200 m

FLLInt-Lane.

= mgLL-Lane*LLInt-Lane. =


c.g.Wheel


= (Wt.-Mid*4.300+Wt.-Fornt*(2*4.300))/(2*145.000+35.000)


dMid-Wheel

Sketch Diagram of Factored Wheel Loads for Exterior Girder :

Sketch Diagram of Girder with Factored Uniformly Distributed & Concentrated DL & LL, Different Locations for Shear & Moments Including Max. Reactions at Supports due to Different DL, LL-Lane Load & LL-Wheel Load :



3L/8 0.728 m L /4 9.150 2.845 4.300

0.300 L /8 6.100 m m 0.300

m 3.050 m m

A B

37.543 kN/m

0.375 m 6.515 kN/m 0.375

m

24.400 m

25.000 m

469.282 kN 490.597 kN

21.316 kN 21.316 kN

81.438 kN 81.438 kN


ii) Calculation of Factored Shear Forces & Moments at Different Locations of Exterior Girder Against AppliedFactored Loads :


Table-6-ii-a-1. Factored Shearing Forces due to all Uniformly Distributed Dead Loads on Exterior Girder.


Table-6-ii-a-2. Factored Moments due to all Dead Loads (DL) on Exterior Girder.



Table-6-ii-b-1. Factored Shearing Forces due to Concentrated Dead Loads of X-Girder on Exterior Girder.

Location From Support-A On Support 0.375m L/8 L/4 3L/8 c.g. L/2 Shearing Forces in kN 12.790 4.263 4.263 -4.263 -4.263 -12.790 -12.790

Table-6-ii-b-2. Factored Moments due to all Dead Loads (DL) on Exterior Girder.



å FDLExt-All =


LSpan =

LTotal =

RA-DL = RB-DL =


RA-LL-L = RB-LL-L =





å FDLExt-All = 41863kN/m

å FLLExt-Lane-Ped. = 65151kN/m


Table-6-ii-c-1. Factored Shearing Forces due to all Applied Dead Loads on Exterior Girder.



Concentrated Loads in kN 12.790 4.263 4.263 -4.263 -4.263 -12.790 -12.790


Table-6-ii-c-2. Factored Moments due to all Applied Dead Loads on Exterior Girder.





d)


Table-6-ii-d-2. Factored Moments on Exterior Girder due to Live Lane Load & Pedestrian Load (LL) .



Table-6-ii-e-1. Factored Shearing Forces on Exterior Girder due to Live Wheel-Load (LL) .



Rear Wheel at 0.375m, AB 81.108 39.421

Rear Wheel at L/8, A B 70.865 29.177

Rear Wheel at L/4, A B 59.185 17.497

Rear Wheel at 3L/8, A B 47.505 5.818


Rear Wheel at L/2, A B 35.826 -5.862


Rear Wheel at L/2, BA 36.768 -56.669

Rear Wheel at 3L/8, B A 69.292 -24.146

Rear Wheel at L/4, B A 69.878 -13.497




Live Load & Pedestrian Loads (åFLLExt-Lane-Ped) :

Table-6-ii-d-1. Factored Shearing Forces on Exterior Girder due to Live Lane Load & Pedestrian Load (LL) .

(FLLExt-Wheel) :

Rear Wheel at L/8, B A 36.477 -5.211

Rear Wheel at 0.375m, B A 41.047 -0.641


Table-6-ii-e-2. Factored Moments on Exterior Girder due to Live Wheel-Load (LL) .




Rear Wheel at 0.375m, A B 81.108 30.416

Rear Wheel at L/8, A B 70.865 216.137

Rear Wheel at L/4, A B 59.185 361.028

Rear Wheel at 3L/8, A B 47.505 434.673


Rear Wheel at L/2, A B 35.826 437.072


Rear Wheel at L/2, B A 36.768 182.778

Rear Wheel at 3L/8, B A 69.292 368.225

Rear Wheel at L/4, B A 69.878 246.998

Rear Wheel at L/8, B A 36.477 111.254

Rear Wheel at 0.375m, B A 41.047 15.393



Table-6-ii-f-1. Sum. of Factored Max. Shear Forces Against All Applied Loads (DL & LL) on Exterior Girder.


470.808 448.204 347.777 224.746 110.242 -40.109 -12.790

79.483 77.040 59.612 39.742 19.871 -4.741 0.000

40.857 39.421 29.177 17.497 5.818 -7.818 -5.862


e)(DL), Live Loads (LL) for Lane Load & Wheel Load & their Summation for Each Point of Application :

Table-6-ii-e-1. Sum. of Factored Max. Moments Against All Applied Loads (DL & LL) on Exterior Girder.


0.000 178.137 1295.697 2242.155 2813.368 3030.496 3035.342

0.000 30.081 218.081 375.557 472.427 508.389 508.691

0.000 30.416 216.137 361.028 434.673 460.734 437.072


7 Factored Shear & Moments at different Positions of an Interior Girder due to Factored Loads (DL& LL) from

Table showing Max. Shear Forces at Different Locations of Exterior Girder due to respective Factored Dead Loads




Table showing the Max. Moments at Different Locations of Exterior Girder due to respective Factored Dead Loads




Superstructure Components & Attachments :

i) Arrangement of Wheel Loads for Interiod Girder & c.g Point :



2.845 m 0.728 m

Rear Middle Front

4.300 4.300


c)

0.728 m

d) 0.728

108.750 kN 108.750 kN 26.250 kN 2.845 c.g.

Rear Middle Front

4.300 4.300

e)


0.300 6.100 m 6.100 m 6.100 m 6.100 m 0.300 m m


12.200 m 3L/8 0.728 m

L /4 9.150 1.455 4.300

0.300 L /8 6.100 m 0.300

m 3.050 m m


c.g.Wheel


= (Wt.-Mid*4.300+Wt.-Fornt*(2*4.300))/(2*145.000+35.000)


dMid-Wheel

Sketch Diagram Showing Factored Wheel Loads for Interior Girder :

Sketch Diagram of Girder with Uniformly Distributed Factored DL & LL, Different Locations for Shear & MomentsIncluding Max. Reactions at Supports due to DL, LL-Lane Load & LL-Wheel Load :


Midd. Wheel Positions under c.g Provisions.Front Wheel Positions under c.g Provisions.

A B

28.710 kN/m

0.375 m 6.200 kN/m 0.375

m

24.400 m

25.000 m

358.875 kN 358.875 kN

42.632 kN 42.632 kN

77.500 kN 77.500 kN


ii) Calculation of Factored Shear Forces & Moments at Different Locations of Interior Girder Against AppliedFactored Loads :


Table-7-ii-a-1. Factored Shearing Forces due to all Uniformly Distributed Dead Loads on Interior Girder.


Table-7-ii-a-2. Factored Moments due to all Dead Loads (DL) on Interior Girder.



Table-7-ii-b-1. Factored Shearing Forces due to Concentrated Dead Loads of X-Girder on Interior Girder.


Table-7-ii-b-2. Factored Moments due to all Dead Loads (DL) on Interior Girder.



å FDLInt-All =

å FLLInt-Lane-Ped =

LSpan =

LTotal =

RA-DL = RB-DL =


RA-LL-L = RB-LL-L =



Girders (å FDLInt-All & FDLInt-Gir-X-Gir.):




å FDLInt-All = 28.710kN/m

å FLLInt-Lane-Ped. = 6.200kN/m

Table-7-ii-c-1. Factored Shearing Forces due to all Applied Dead Loads on Interior Girder.



Concentrated Loads in kN 25.579 25.579 25.579 8.526 8.526 -8.526 -8.526


Table-7-ii-c-2. Factored Moments due to all Applied Dead Loads on Interior Girder.





d)


Table-7-ii-d-2. Factored Moments on Interior Girder due to Live Lane Load & Pedestrian Load (LL) .



Table-7-ii-e-1. Factored Shearing Forces on Interior Girder due to Live Wheel-Load (LL) .



Rear Wheel at 0.375m, AB 211.587 102.837

Rear Wheel at L/8, A B 184.864 76.114

Rear Wheel at L/4, A B 154.395 45.645

Rear Wheel at 3L/8, A B 123.927 15.177


Rear Wheel at L/2, A B 93.458 -15.292


Rear Wheel at L/2, BA 150.292 -93.458

Rear Wheel at 3L/8, B A 180.761 -62.989

Rear Wheel at L/4, B A 182.290 -35.210

Rear Wheel at L/8, B A 95.156 -13.594

Rear Wheel at 0.375m, B A 107.079 -1.671



Live Load & Pedestrian Loads (åFLLInt-Lane-Ped) :

Table-7-ii-d-1. Factored Shearing Forces on Interior Girder due to Live Lane Load & Pedestrian Load (LL) .

(FLLInt-Wheel) :

Table-7-ii-e-2. Factored Moments on Interior Girder due to Live Wheel-Load (LL) .




Rear Wheel at 0.375m, A B 211.587 79.345

Rear Wheel at L/8, A B 184.864 563.836

Rear Wheel at L/4, A B 154.395 941.812

Rear Wheel at 3L/8, A B 123.927 1133.930


Rear Wheel at L/2, A B 93.458 1140.188


Rear Wheel at L/2, B A 150.292 1140.188

Rear Wheel at 3L/8, B A 180.761 960.586

Rear Wheel at L/4, B A 182.290 644.344

Rear Wheel at L/8, B A 95.156 290.227

Rear Wheel at 0.375m, B A 107.079 40.154



Table-7-ii-f-1. Sum. of Factored Max. Shear Forces Against All Applied Loads (DL & LL) on Interior Girder.


375.841 365.075 288.276 183.657 96.092 -29.418 -8.526

75.640 73.315 56.730 37.820 18.910 -4.512 0.000

106.583 102.837 76.114 45.645 15.177 -20.394 -15.292


e)(DL), Live Loads (LL) for Lane Load & Wheel Load & their Summation for Each Point of Application :

Table-7-ii-e-1. Sum. of Factored Max. Moments Against All Applied Loads (DL & LL) on Interior Girder.


0.000 142.152 1039.048 1811.021 2263.908 2442.182 2449.720

0.000 28.627 207.537 357.399 449.585 483.808 484.096

0.000 79.345 563.836 941.812 1133.930 1201.915 1140.188


Table showing Max. Shear Forces at Different Locations of Interior Girder due to respective Factored Dead Loads




Table showing the Max. Moments at Different Locations of Interior Girder due to respective Factored Dead Loads




N/mm/mm-Wd.


H. Strength Limit State Design (USD) of Main Girder & Cross Girders Against Applied Forces :

1 Data for Flexural Design :


i) Dimensional Data of Superstructure :




d) Thickness of Deck Slab 0.200 m

e) Thickness of Wearing Course 0.075 m

f) Number of Main Girders 5 nos

g) Number of Cross Girders 5 nos

h) Depth of Main Girders (Including Slab as T-Girder) 2.000 m

i) Depth of Cross Girders (Including Slab as T-Girder) 1.900 m

j) Width of Web for Main Girders 0.350 m

k) Width of Web for Cross Girders 0.250 m

l) C/C Distance Between Main Girders 2.000 m

m) Distance of Slab Outer Edge to Exterior Girder Center 1.125 m

n) Clear Distance Between Main Interior Girders 1.650 m

o) Filets : i) Main Girder in Vertical Direction 0.150 m

p) ii) Main Girder in Horizontal Direction 0.150 m



2 Design Criterion, Loadings, Design Data (Materials) & Different Factors :

i) Design Criterion : AASHTO Load Resistance Factor Design (LRFD-USD).

ii) Type of Loads : Combined Application of AASHTO HS20 Truck, Lane & Pedestrian Loadings.

iii) Design AASHTO HS20 Truck Loading :









SL

SAddl.

LGir.

hSlab.

hWC

NGir.

NX-Gir.

hGir.

hX-Gir.

bWeb-Gir.

bWeb-X-Gir.

C/CD-Gir.

CD-Ext.-Gir-Edg.

ClD-Int.-Gir.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

DAxel.

DWheel.

LLRW-Load

LLRSW-Load

LLMW-Load

LLMSW-Load

LLFW-Load

LLFSW-Load


iv) Design AASHTO Lane Loading :

a) Design Lane Loading is an Uniformly Distributed Load having Magnitude of 9.300 N/mm 9.300N/mm through the Length of Gridge for Single and acting over a 3.000m 9.300 kN/m Wide Strip in Transverse Direction.

v) Design AASHTO Pedestrian Loading :

a) Design Pedestrian Loading is an Uniformly Distributed Load having Magnitude 0.003600 N/mm^2

3.600 kN/m^2 the total Wide of Sidewalk.

vi)

9.807






vii)

a) Unit weight of Normal Concrete 24.000 kN/m^3

b) Unit weight of Wearing Course 23.000 kN/m^3

c) Unit weight of Normal Water 10.000 kN/m^3

d) Unit weight of Saline Water 10.250 kN/m^3

e) Unit weight of Earth (Compected Clay/Sand/Silt) 18.000 kN/m^3


a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) Poisson's Ration = 0.200, subject to cracking and considered to be neglected 0.200

e) 2.887 MPa

f) 410.000 MPa

g) 164.00 MPa

h) 200000 MPa

ix) Conventional Resistance Factors for Ultimate Stressed Design & Construction (AASHTO LRFD-5.5.4.2.1) :

LL-Lane

LL-Pedest

of 3.600*10-3MPa through the Length of Sidewalk on both side and acting over



gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc

wWC

wWater-Nor.

wWater-Sali.

wEatrh


c



c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).












d) For Axil Comression with Spirals or Ties & Seismic Zones at Extreme Limit 0.75 State (Zone 3 & 4).






j) 0.85 (AASHTO LRFD-5.7.2..2.)

h) 0.85

viii) Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :








ix) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


b) 1.750

c) IM 1.330 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1;

fFlx-Rin.

fFlx-Pres.

fShear/Torsion.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.





gEH


gEV


gES





(Applicable only for Truck Loading & Tandem Loading)

d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

x) Design Data for Site Conditions :


b) Velocity of Wind Load in Cyclonic Storm Condition 260.000 km/hr



















VWL-Nor.

VWL-Spe.

VWA


3 Design Phenomena, Selection of T-Girder Flange Width, Girder Depth & Calculations for Monent & Shear :

i) Design Phenomena :

a)

b)

respective Moments & Shears. Since the Bridge Deck Slab is integral Part of Girders, thus the Design of Girders

c)a reasonable Steel Area, thus an Equivalent Additional Steel Area can also Provide on Tension Face against those Shrinkage & Temperature Reinforcement. These arrangement will provide a Higher Stiffness for the Structure within Flexural provisions.

ii) Cross Sectional Sketch Diagram of Bridge Girders & Dack Slab : 1.475

0.225

7.300 1.250

0.300

1.070

0.300

0.200

0.250

0.950 1.650 1.650 1.650 1.650 0.950

1.125 2.000 2.000 2.000 2.000 1.125

10.250

iii) Selection of T-Girder Effective Flange Width under Provisions of AASHTO-LRFD-4.6.2.6 (4.6.2.6.1) :

a) Since the Bridge is a Simple Supported T-Girder Structure, thus Falnge Width will be the least Dimention of :

= 6.100 m

2.750 m

= 2.000 m

b) From Calculations, Average Spacing of Adjacent Beams/Girders is the 2.000 m

The Flexural of Girders will be according to AASHTO LRFD or Ultimate Strength Design (USD) Procedures.

Since the Interior Girders of the Bridge have the Max. Moments & Shearing Forces caused by Applied Loads (DL& LL), thus it is require to conduct the Flexural Design for Reinforcements of Bridge Girders Based on Calculated

will be under T-Beam if the Provisions in these Respect Satisfy, otherwise Designee will be under Provisions for theRectangular Beam.

The T-Girder will have to Provide Longitudinal Shrinkage & Temperature Reinforcement on Compression Face with

i) One-quarter of Effective Span Length = 1/4*SL

ii) 12.0 times average Depth of Slab + Greater Thickness of Web = 12*hSlab + bGir. =iii) One-half the Width of Girder Top Flange (It is not req. as there is no Addl. Top Flange)

iv) The average Spacing of Adjacent Beams/Girders = C/CD-Gir.

bFl-Gir.

Least one, thus the Flange Width of Interior Girders, bFla-Gir = 2.000m

CL


iv) Selection of Depth (Including Slab Depth) for T-Girder under Provisions of AASHTO-LRFD-2.5.2.6.3 & Table2.5.2.6.3-1 :

a)

b) L 24.400 m

c) 1.732 m

d) Considering the Clear Covering both on Top & Bottom, Let Provide a Depth for 2.000 mthe T-Girder = 2.000m.

v) Calculations for Monent at Different Location of Girder :

a) From Load, Shear & Moment Calcutation Tables it appares that, the Interior Girders are facing the Max. Resultant

b)

Table-1. Sum. of Max. Moments Against All Applied Loads (DL & LL) on Interior Girder.


0.000 181.672 1,327.681 2,313.495 2,892.428 3,120.031 3,129.494

0.000 50.096 363.190 625.448 786.774 945.861 847.168

0.000 184.676 1,312.328 2,192.069 2,639.221 2,797.457 2,653.786

Total Moments on Each Point 0.000 278.135 3,003.199 5,131.011 6,318.423 6,863.350 6,630.449

c) Moment at Sopport Position of Girder 0.000 kN-m

d) 278.135 kN-m

e) 3,003.199 kN-m

f) 5,131.011 kN-m

g) 6,318.423 kN-m

h) 6,863.350 kN-m

i) 6,630.449 kN-m

j)

Since the Bridge is a Simple Supported T-Girder Structure, thus According to Table-2.5.2.6.3-1; the Min. Required Girder Depth Including Salb Thickness is = 0.0708L; where

L is the Span Length, the Clear Distance between Bearing Centers of Supports

= SL

Thus required Minimum Depth of T-Girder Including Salb = 0.070*SL hT-Girder.

hGir-pro.

Forces (DL & LL) causing Max. Shears & Moments, thus One of Interior Girders is considered as Typical one for theFlexural Design in respect of All Applied Loads (DL & LL) and Corresponding Moments & Shears.

Table for Max. Moments at Different Locations of Interior Girder due to Factored DL, Lane-LL & Wheel-LL :

a. Dead Load (åFDLInt)

b. Lane Live Load (åFLLInt)

c. Wheel Live Load (WLLInt)

Mu-Support.

Moment at a Distance 0.375m from Sopport of Girder Mu-0.375m.

Moment at a Distance L/8 from Sopport of Girder Mu-L/8.


Moment at a Distance 3L/8 from Sopport of Girder Mu-3L/4.

Moment at Absolute Max. Moment Loaction (c.g. Position) of Girder Mu-c.g.

Moment at a Distance L/2 (Middle of Span) from Sopport of Girder Mu-L/2.

Sketch Diagram of Main Girder T-Beam :


b2.000

0.200 m

2.000 m1.808 m

0.350 m

vi) Limits For Maximum Reinforcement, (AASHTO-LRFD-5.7.3.3.1) :.

a) With Maximum Amount of Prestressed & Nonprestressed Reinforcement for a 0.420

b) c Variable

c) Variable

Variable

Variable

Variable

Variable mm

Variable mm


vii) Limits For Manimum Reinforcement, (AASHTO-LRFD-5.7.3.3.2) :



-

hf =

hGir. =d =

bWeb =

c/de-Max.











Centroid in mm.




thus de = ds .






at Extreme Fiber where Tensile Stress is caused by Externally Applied Forces

2,449.720 N-mm

Variable

233333333

0.233333

233.333/10^3

2.887

c) 673638627.159 N-mm

673.639 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)

Location of Value of Value of Actuat Acceptable M Maximum

Section Unfactored Cracking Factored Allowable Flexuralfrom Dead Load As per Moment Cracking Cracking Moment Factored Min. Moment Moment

Support Moment Equation Value Moment Moment of Section Moment

5.7.3.3.2-1 M (1.33*M)kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

At Support 0.000 673.639 673.639 673.639 808.366 0.000 0.000 808.366 808.366

142.152 673.639 673.639 673.639 808.366 278.135 369.920 808.366 808.366

1039.048 673.639 673.639 673.639 808.366 3003.199 3994.255 808.366 3003.199

1811.021 673.639 673.639 673.639 808.366 5131.011 6824.245 808.366 5131.011

2263.908 673.639 673.639 673.639 808.366 6318.423 8403.503 808.366 6318.423

2449.720 673.639 673.639 673.639 808.366 6863.350 9128.255 808.366 6863.350

viii) Flexural Design of Main Girder with Max, Moment value (At c.g. Position) :

a) 6,863.350 kN-M

6863.350*10^6 N-mm, thus 6863.350*10^6 N-mmthis value is Considerd as the Moment at Middle Position of Span.

after allowance for all Prestressing Losses in MPa. For Nonprestressing RCC

Components value of fcpe = 0.







For the Rectangular RCC Girder Section value of Snc = (bWebhGir3/12)/(hGit/2) m3







Table-3 Showing Allowable Resistance Moment M r for requirment of Minimum Reinforcement at Different Sections



for RCC Mu


At L0.375m

At L/8At L/4At 3L/8At c.g/L/2

The Absolute Max. Moments on Interior Girder is at c.g. Point. Since it is very MU

close to Middle Position of Span having value MU.=


b) 50 mm

50 mm

38 mm

c) 32 mm

d) 804.248

e) 32 mm

f) 12 mm

g) Let Assume Main Tensile Reinforcements are being placed in 4 Layers each 1,826.000 mmhaving Equal nos. of Bars, thus the Effective Depth of Reinforcements from Top of Girder up to the Centroid of Assumed Group of Reinforcements

h) 0.022

i) 0.0165

ix) Checking's Whether the Bridge Girder would Designed as T-Beam or Rectangular Beam Provisions :

a) According to Ultimate Stressed Design Provisions a Rectangular having Flange with Reasonable Thickness on its

b)Rectangular Beam.

c) a 108.509 mm

a<hFln

d)Flexural Design of T-Girder will as a Rectangular Beam Provisions.

4 Flexural Design for Provisions of Tensile Reinforcements at Different Section of Bridge Girder :

i) Provision of Tensile Reinforcements at Central Section at L/2 from Support (Mid Span) against CalculatedMoments :

a) The Calculated Factored Max. Moment at Central Section is being Considered 6863.350 kN-mthe Factored Moment at Absolute Moment Position which is also Greater than 6863.650*10^6 N-m

808.366 kN-m

Let the Clear Cover at Bottom Surface of Girder, C-Cov.Bot. = 50mm, C-Cov-Bot.

Let the Clear Cover at Top of Girder, C-Cov.Top = 50mm, C-Cov-Top.

Let the Clear Cover at Vertical Faces of Girder, C-Cov.Vert. =38mm, C-Cov-Side.

Let the Main Reinforcements are 32f Bars in 4 Layers, DBar

X-Sectional Area of Main Reinforcements Af = p*DBar2/4mm2 Af-32 mm2

The Vertical Spacing between Reinforcement Bars, sVer. = 32 mm sVer.

Let the Transverse/Shear Reinforcements (Stirrups) are of 12f Bars, DStir.

dasu-L/2

= (hGir - C-Cov-Bot -DStri -2*DBar - 1.5*sVer.)

Balanced Steel Ratio for Grider Section according to AASHTO-1996-8.16.2.2 rb.

rb. = (0.85*0.85*(f/c/fy)*(599.843/(599.843+fy))

Max. Steel Ratio, rMax = 0.75*rb. (AASHTO-1996-8.16.2.1) rMax

Top should be Designed as T-Beam if Depth of Equivalent Compression Block 'a' is Less than Flange Thickness

Let Consider the T-Girder will behave as Rectangular Beam for which the Total Flange Width-'b' will be the Width of

For a Rectangular Section having Assumed Effective Depth dacu; Beam Width

b and the Calculated Max. Factored Ultimate Moment (Absolute Moment) MU;

the value Equivalent Compression Block a = de(1 - (1 - 2MU/0.85f/cbde

2)1/2)

Since the Calculated value of Equivalent Compression Block a < hFln; the thickness of Girder Flange, thus

MCent.

the Required Minimum Moment Mr, thus MCent. is the Governing Moment in Mr.


Flexural Designe for the Central (Mid) Section of Girder. 808.366*10^6 N-m

b) 6,863.35 kN-m

6863.650*10^6N-m

c) 10,498.045

against Factored Moment for the Section, the Required Tensile Steel Area for

d) 13.053 nos.

e) 18 nosthe Bottom 4-Layers & 2 nos on Top Layer.

f) 14,476.459

g) With provided arrangement of Reinforcement Bars & Steel Area for the Section 1,808.222 mmthe Actual Effective Depth of Tensile Reinforcement's Centroid from the Extreme

h) 166.256 mm

a<hFln Satisfied

i) The Developed Resisting Moment against provided Steel Area for the Section, 10239.034 kN-m

Mr>Mu Satisfied

j) 0.004 p-pro<p-max Satisfied

k)

ii) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

a) 0.420

b) c 141.318 mm

c) 0.85

Since the Calculated Factored Moment MCent. is the Governing Moment for the MU

Flexural Design, thus it is also the Uiltimate Design Moment MU.

With MU, Design Moment; b, Width of Rectangular Beam; dasu-L/2, Assumed As-req.-L/2 mm2

Effective Depth for the Section & 'a' Calculated Equivalent Compression Block

the Section; As = MU/[ffy(dasu-L/2 - a/2)]

Number of 32f bers required = As-req-Total/Af-32 NBar-req

Let Provide 18nos 32f bars in 5 (Five) Layers having 4 nos. of Bars on each of NBar-pro.

Provided Steel Area for the Section with 20 nos. 32f bars = Nbar-pro*Af As-pro mm2

de-pro

Fiber of Compression Face = {4*Asf-32*(hGir - CCov-Bot -DStir.-DBar/2)

+ 4*Asf-32(hGir-CCov-Botr-DStir.-DBar-sVer.-DBar/2) +4*Asf-32(hGir-CCov-Bot-DStir.-2*DBar-2*sVer.-DBar/2)

+ 4*Asf-32(hGir-CCov-Bot-DStir.-3*DBar-3*sVer.-DBar/2)+2*Asf-32(hGir-CCov-Bot-DStir.-4*DBar-4*sVer.-DBar/2)}/As-pro

Value of Equivalent Compression Block 'a' against Provided Steel Area for apro

Rectangular Section of Girder = As-pro*fy/(0.85*f/c*b)

MResis

= As-pro*fy(de-pro - apro/2)/106

Steel Ratio against Provided Steel Area for the GirderSection = As-pro./b*de-pro rpro

Since against Provided Steel Area; i) the Equivalent Compression Block 'a'< hFln, Thickness of Girder Flange; ii) the

Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. AllowableMax. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Central Section is OK.


c is the Distance between Neutral Axis & Extrime Compressive Face, having

value of c = b1apro, in mm.



d) 0.078

f) c/de-pro<c/de-max. OK

iii) Provision of Tensile Reinforcements at Section 3L/4 from Support against Calculated Moments:

a) 6318.423 kN-m

6318.423*10^6 N-m

808.366 kN-m808.366*10^6 N-m

b) 6318.423 kN-m

6318.423*10^6 N-m

c) 9,664.537

d) 12.017 nos.

e) 20 nosthe Bottom 4-Layers & 2 nos on Top Layer.

f) 16,084.954


h) 184.729 mm

a<hFln Satisfied


Mr>Mu Satisfied


k)



The Calculated Factored Moment a Section 3L/8 from Support is being Greater M3L/8.

than the Required Minimum Moment Mr, thus the Calculated Factored Moment.

M3L/8. is the Governing Moment in Flexural Designe for the Section 3L/8.

Since the Calculated Factored Moment M3L/8. is the Governing Moment for the MU


With MU, Design Moment; b, Width of Rectangular Beam; dasu-L/2, Assumed As-req.-3L/8 mm2

Effective Depth for the Section & 'a' Calculated Equivalent Compression Blockagainst Factored Moment for the Section L/2, the Required Tensile Steel Area

for the Section at 3L/8; As = MU/[ffy(dasu-L/2 - a/2)]


Let Provide 18nos 32f bars in 5 (Five) Layers having 4 nos. of Bars on each of NBar-pro.

Provided Steel Area for the Section with 16nos. 32f bars = Nbar-pro*Af As-pro mm2

de-pro



+ 4*Asf-32(hGir-CCov-Bot-DStir.-3*DBar-3*sVer.-DBar/2)+2*Asf-32(hGir-CCov-Bot-DStir.-4*DBar-4*sVer.-DBar/2)}/As-pro



MResis




Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. Allowable


iv) Provision of Tensile Reinforcements at Central Section at L/4 from Support against Calculated Moments :

a) 5131.011 kN-m

5131.011*10^6 N-m

808.366 kN-m 808.366*10^6 N-m

b) 5,131.011 kN-m

5131.011*10^6 N-m

c) 7,848.29

d) 9.759 nos.

e) 14 nosBottom 3-Layers & 2 nos on Top Layer.

f) 11,259.468


h) 21.266 mm

a<hFln Satisfied


Mr>Mu Satisfied


k)

v) Provision of Tensile Reinforcements at Section L/8 from Support against Calculated Moments :

Max. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Section at 3L/8 is OK.

The Calculated Factored Moment a Section L/4 from Support is being Greater ML/4.

than the Required Minimum Moment Mr, thus the Calculated Factored Moment.

ML/4. is the Governing Moment in Flexural Designe for the Section L/4.





for the Section at L/4; As = MU/[ffy(dasu-L/2 - a/2)]


Let Provide 14nos 32f bars in 4(Four) Layers having 4 nos. of Bars on each of NBar-pro.


de-pro



+ 2*Asf-32(hGir-CCov-Bot-DStir.-3*DBar-3*sVer.-DBar/2))}/As-pro



MResis




Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. AllowableMax. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Section at L/4 is OK.


a) 3003.199 kN-m

3003.199*10^6 N-m

808.366 kN-m 808.366*10^6 N-m

b) 3,003.199 kN-m

3003.199*10^6 N-m

c) 4,593.635

d) 5.712 nos.

e) 10 nosof the Bottom 2-Layers & 2 nos on Top Layer.

f) 8,042.477


h) 92.365 mm

a<hFln Satisfied


Mr>Mu Satisfied


k)

vi) Provision of Tensile Reinforcements at Section at 0.375m from Support against Calculated Moments :

a) 278.135 kN-m

278.135*10^6 N-m

The Calculated Factored Moment a Section L/8 from Support is being Less ML/8

than the Required Minimum Moment Mr, thus the Required Minimum Momen

Mr. is the Governing Moment in Flexural Designe for the Section L/8.





for the Section at L/8 ; As = MU/[ffy(dasu-L/2 - a/2)]


Let Provide 10 nos 32f bars in 3 (Three) Layers having 4 nos. of Bars on each NBar-pro.


de-pro


+ 4*Asf-32(hGir-CCov-Botr-DStir.-DBar-sVer.-DBar/2)

+ 2*Asf-32(hGir-CCov-Bot-DStir.-2*DBar-2*sVer.-DBar/2))}/As-pro



MResis




Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. AllowableMax. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Section at L/8 is OK.

The Calculated Factored Moment a Section 0.375m from Support is being Less M0.375m

than the Required Minimum Moment Mr, thus the Required Minimum Momen


808.366 kN-m808.366*10^6 N-m

b) 808.366 kN-m

808.366*10^6 N-m

c) 1236.461

d) 1.537 nos.

e) 8 nosthe Layers.

f) 6,433.982


h) 73.892 mm

a<hFln Satisfied


Mr>Mu Satisfied


k)

vii) Provision of Tensile Reinforcements at Section on Support against Calculated Moments :

a) 0.000 kN-m

000*10^6 N-m808.366 kN-m

808.366*10^6 N-m

b) 808.37 kN-m

Mr is the Governing Moment in Flexural Designe for the Section 0.375m.

Since the Required Minimum Momen Mr. is the Governing Moment for the MU


With MU, Design Moment; b, Width of Rectangular Beam; dasu-L/2, Assumed As-req.-0.375 mm2


for the Section at 0.375m ; As = MU/[ffy(dasu-L/2 - a/2)]


Let Provide 8nos 32f bars in 2 (Two) Layers having 4 nos. of Bars on each of NBar-pro.

Provided Steel Area for the Section with 8 nos. 32f bars = Nbar-pro*Af As-pro mm2

de-pro


+4*Asf-32(hGir-CCov-Bot-DStir.-*DBar-*sVer.-DBar/2))}/As-pro



MResis




Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. AllowableMax. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Section at 0.375m is OK.

The Calculated Factored Moment at Support Position is being 0 (Zero), thus MSupp.

for Support Section the Required Minimum Moment Mr, is the Governing Governing Moment in Flexural Designe for the Section.

Since the Required Minimum Momen Mr. is the Governing Moment for the MU


808.366*10^6 N-m

c) 1,236.46

d) 1.537 nos.

e) 8 nosthe Layers.

f) 6,433.982


h) 73.892 mm

a<hFln Satisfied


Mr>Mu Satisfied


k)

viii) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 9,132.190 kN-m 91132.1902*10^6 N-mm

10,146.878 kN-m 10146.878*10^6 N-mm

f 0.90

b)


With MU, Design Moment; b, Width of Rectangular Beam; dasu-L/2, Assumed As-req.-Supp. mm2


for the Section at Support ; As = MU/[ffy(dasu-L/2 - a/2)]


Let Provide 8nos 32f bars in 2 (Two) Layers having 4 nos. of Bars on each of NBar-pro.


de-pro


+4*Asf-32(hGir-CCov-Bot-DStir.-*DBar-*sVer.-DBar/2))}/As-pro



MResis




Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. AllowableMax. Steel Ratio; thus the Provision & Flexural Design for Tensile Reinforcement for Section at Support is OK.




The Nominal Resistance for a Flanged Section with One Axis Stress having both Prestressing & Nonprestessing


y(d/s-a/2) + 0.85f/



c)

d) 10,146.878 kN-mSteel Area against Factored Max. Moments at its Mid Span will have value of 10146.878*10^6 N-mm

e) 6,863.350 kN-m6863.350*10^6 N-mm

f) Mr>Mu

Satisfied

ix) Checking in respect of Control of Cracking By Distribution of Reinforcement, (AASHTO-LRFD-5.7.3.4) :

a)

Where;

b) 157.694

4,127.905 kN-m 4127.905*10^6 N-mm

14,476.459

1,808.222 mm Tensile Reinforcement for the Section.

c) 291.299

66.000 mmTension Bar. The Depth is Summation Bottom/Top Clear Cover & Radius of the

A 7,458.025 by Dividing the Total Concrete Area bounded in between Extreme Tension Face & a Straight Line parallel to Neutral Axis of Component having equal distance fromthe Centrioed of Main Tension Reinforcement Bars on both side & Diving the Area by the total Number of Main Bars as Tensile Reinforcement having 50mm.Max.

For a Nonprestressing Structural Component having Either of I or T Section with Flenge & Web Elements, at any

Section the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

For Simple Supported & Single Reinforced T-Girder Beam the with Provided Mn-Mid-Span

Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

Calculated Factored Moment MU at Absulate Max. Moment at c.g. of MU-Mid-Span-USD

T-Girder Span).



Under Service Limit State Load Condition, Developed Tensile Stress of Reinforcement fs-Dev. of a Concrete Elements,

should not exceed fsa the Computed Tensile Stress of Reinforcement under provision of AASHTO-LRFD-5.7.3.4.

fs-Dev. is Developed Tinsel Stress in Provided Reinforcements of Section under fs-Dev. N/mm2

the Service Limit State of Loads = M/As-prode in which,

i) M is Calculated Moment for the Section under Service Limit State of Loads MWSD.


iii) de is Effective Depth between Extreme Compression Fiber to Centroid of the de



i) dc = Depth of Concrete Extreme Tension Face from the Center of the Closest dc

Closest Bar to Tension Face. The Max. Clear Cover = 50mm. For a Section of

T-Girder Component, dc = DBar/2 + CCov-Bot. Since Clear Cover at Bottom of

T-Girder, CCover-Bot = 50mm & Bar Dia, DBar = 32f ; thus dc = (32/2 + 50)mm



Clear Cover. The Girder is being Provided with 50mm Clear Cover at Bottom, 18nos.

of Tension Reinforcement are 32mm but due to dissimilarity of Bar arrangement the Centroid of Bars does not Coincide with the Center of Vertical Spacings.

23,000.000 N/mm


246.000

d)

e) 12,451.016 N/mm

f) fs-Dev.< fsa Satisfied

g) fsa > 0.6fy Not Satisfied


i)

j)

x)

(Absulate Max. Moment at c.g. Point); v) The Provided Main Reinforcements are anough in respect of Control of Cracking by Distribution of Reinforcement;

5 Design of Longitudinal Reinforcement on Top & Vertical Surfaces of Girder :

i) Shrinkage & Temperature Reinforcement in Longitudinal Direction on Top Surface of Girder :

32f Bars Tension Bar in 5 Layers on Mid Span having 4 nos. Bars on Botton 4 Layers & 2nos. on Top Layer. Though the Vertical Spacing among the Layers

Thus, based on actual Cintroid of Bars value of A = bWeb{2*(hGir. - de-pro)}/NBar-pro.





Since the Calculated value of fs-Dev. is responsible for Controlling the formation of Cracks under Applied Loads to the

T-Girder Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.





Since the Developed Tensile Stress of Tension Reinforcement of T-Girder fs-Dev.< fsa the Computed Tensile Stress

with Alloable Max. Crack Width Parimeter ZMax.; the Developed Crack Width Parimeter ZDev. < ZMax. thus Provision of Main Tensial Reinforcement for T-Girder in respect of Control of Cracking by Distribution of Reinforcement are OK

More over though the Structure is a Nonprestrssed one, but value of dc have not Exceeds 900 mm, thus Component does require any additional Longitudinal Skein Reinforcement.

Since, i) The Value of Resisting Moment > Design Moment;

ii) The Calculated value of c/de £ 0.42;

iii) The Provided Stee Ratio for Rectangular Section of T-Girder, rRec-pro < rMax. Balance Steel Ratio;

iv) The Computed Factored Flexural Resistance Mr > Mu the Actual Factored Moment at Mid Span

Thus Flexural Design of Reinforcements for the T-Girder Span is OK.


a) Since the Girders are Simply Supported Structure & at Middle Location the Moments are with (+) ve. value, thus the Top Surface of Girder will be under Compression. Under Single Reinforced Flexural Design Provision Top Surface ofGirder does not require any Longitudinal Reinforcement. Yet for Safe Guard & provide the Web/Shear Reinforcements in the form of Stirrups, this Surface is also require Longitudinal Reinforcement, which can arrange under the Provision

b) Let consider 1 (One) meter Strip Length of Girder for Calculation of Shrinkage 1.000 m& Temperature Reinforcement in Longitudinal Direction on Top Surface of Girder, 1,000.000 mm

c) Let consider the Width of Girder is the Web Width of T-Girder = 0.450m 0.350 m 350.000 mm

d) 93.902 Temperature Reinforcement for Structural Components with Thickness Less

e) 350000.000

f) 32.000 mm

g) 804.248

h) 2.000 nos.Surface of Girder.

i) Provide Steel Area for Longitudinal Shrinkage & Temperature on Top Surface of 1,608.495

j) Spacing of Shrinkage & Temperature Reinforcements will less of 3-times the 450 mm

1,400.00 mm or = 450 mm

k) 186 mm

l) Whether Provision of Longitudinal Shrinkage & Temperature Reinforcement As-pro-S&T>As-req.Tophave Satisfied or not. Satisfied

m) Whether Provision of Spacing have Satisfied or not. s-pro-Top < s-req.TopSatisfied

n)

ii) Shrinkage & Temperature Reinforcement in Longitudinal Direction on Vertical Faces of Girder :

of Shrinkage & Temperature Reinforcement according to AASHTO-LRFD-5.10.8.

LGirder.

bWeb.

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage & As-req-Top. mm2

than 1200mm; As ³ 0.11Ag/fy . (Width of Girder is its thicknes, b = 4500mm < 1200mm).

Here Ag is Gross Area of Girder Top Surface = LGirder*bWeb Ag-Top mm2

Let provide 32f bars as Longitudinal Shrinkage & Temperature on Top Girder. DBar.

X-Sectional Area of 32f bar = pDBar2/4 Af-32 mm2

Let provide 2 nos. 32f bars as Longitudinal Shrinkage & Temperature on Top NBar-S&T-Top

As-pro-S&T. mm2

Girder = NBar-S&T-Top*Af-32

s-req-Top.

Component thickness = 4*bWeb =

Spacing in between 2 nos. 32f Longitudinal bars on Top Surface s-pro-Top.

= (b - 2.*CCover-Side - 2*DStir. - 2*DBar)

Since As-pro-Top. > As-req-Top & spro-Top < s-req-Top, thus the provision of Longitudinal Shrinkage & Temperature on Top Surface of Girde is OK.


a) Girders are Simply Supported Structure having the Longitudinal Flexural Reinforcements due to Moments on Bottom Surface & also on Top Surface (In case of Doubly Reinforced or under Shrinkage & Temperature Provision). There willbe also Lateral & Vertical Reinforcements on Bottom, Top & Vertiocal Surfaces under the provisiona of Shear & WebReinforcement. It also requires Longitudinal Reinforcement on its Vertical Faces, those can provide under Shrinkage

b) Let consider 1 (One) meter Strip Length of Girder for Calculation of Shrinkage 1.000 m& Temperature Reinforcement in Longitudinal Direction on Vertical Faces of Girder. 1,000.000 mm

c) 2.000 m 2,000.000 mm

d) 536.585 & Temperature Reinforcement for Structural Components with Thickness Less

e) 2000000.000

f) 16 mmFaces of Girder.

g) 201.062

h) Spacing of Shrinkage & Temperature Reinforcements is less of 3-times the 450 mm

1,050 mm or = 450 mm

i) 2.669 nos.Shrinkage & Temperature Reinforcements on Vertical Surface of Girder

j) 374.096 mmTemperature Reinforcementon Vertical Faces of Girder for Calculated Steel Area

k) 5 nosReinforcement.

l) 232.857 mmTemperature Reinforcementon Vertical Faces of Girder for Calculated Steel Area

m) Steel Area against Provided Longitudinal Shrinkage & Temperature 1,005.310

n) Whether Provision of Longitudinal Shrinkage & Temperature Reinforcement As-pro-S&T-Hori. > As-req-S&T.-Hori.

& Temperature Reinforcement Provisions according to AASHTO-LRFD-5.10.8.

LGirder.

Let consider the Depth of T-Girder as Height of Girder, h = 2.00m hGirder.

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T.-Hori. mm2

than 1200mm; As ³ 0.11Ag/fy .

Here Ag is Gross Area of Girder's Each Vertical Face = LGirder*hGirder Ag-Vert. mm2

Let provide 16f bars as Longitudinal Shrinkage & Temperature on Vertical DBar.


s-req-S&T.-Hori.


Number of 16f bars required against Calculated Steel Area as Horizontal NBar-req-S&T-Hori.

= As-req-Long./Af-16

Spacing of 16f bars at Support Position as Horizontal Shrinkage & sCal-req-S&T-Hori.

= (hGir -2*Ccov -2.669*DBar-32 - sVer- 2*DStri)/NBar-req-S&T-Hor.

Let Provide 5 nos. 16f as Longitudinal bars as Shrinkage & Temperature NBar-pro-S&T-Hori.

Spacing of 16f bars at Support Position as Horizontal Shrinkage & spro-S&T-Hori.

= (hGir -2*Ccov -5*DBar-32 - sVer- 2*DStri)/NBar-req-S&T-Hor.

As-pro-S&T-Hori. mm2

Reinforcement on Vertical Surfaces of Girder = Af-16*Barpro-S&T-Hori.


have Satisfied or not. Satisfied

o) Whether Provision of Spacing have Satisfied or not. spro-S&T-Hori..<s-req-S&T.-HoriSatisfied

p)

4 Checking for Development Length & Splices of Reinforcements under Provisions of AASHTO-LRFD-5.11 :

i) Provisions Article 5.11.1.1.1 :

a) The Calculated Force effects in Reinforcement at each Section shall be Developed on each side of that Section byEmbedment of Length of Reinforcement, Hooks or any Mechanical Device or a Combination of all together. Theseare especially required for Tension Zones.

ii) Provisions Flexural Reinforcement (Article 5.11.1.2.1) :

a) In Flexural Members the Critical Sections are,i) The Points of Maximum Stress; ii) The Points within the Span where Adjacent Reinforcement Terminates or Bands.

b) Except at Supports of Simple Spans and at the Free Ends of Cantilevers, Reinforcements shall be extended furtherbeyond the Point where they are no longer requred to resist the Flexural for a Distance as mentioned below:i) Not less than the Effective Depth of Component;ii) Not Less than the Nominal Diameter of Bar proposed, andiii) Not Less than 1/20 of Clear Span.

c)

d) In no case Termination of more than 50% of Reinforcement at any Section is permitted & Termination of adjusecentBars on the same Section is restricted.

f)

on the Compression Face of the Component.

g) Components of i) Sloped, Stepped or Tapered Footings; ii) Brackets; iii) Deep Flexural Members; & iv) UnparallelTension & Compression Reinforcements and the Reinfocement Forces are not Directly Proportional to the FactoredMoments, those Provide with Supplementary Ancorages for Flexural Tension Reinforcements.

iii)

a)

Since As-pro-S&T-Hori. > As-req-Long & spro-S&T-Hori.< s-req-S&T-Hori. thus the provisions of Longitudinal Reinforcement as Shrinkage & Temperature on Vertical Faces of Girder is OK.

In Continuing Reinforcements the Extended Length shall not be Less than the Development Length ld, beyond the

Point of Bent or Tarminated Tension Reinforcenenta are no longer requred. The value of Development Length Id willbe according to Provisions as mentioned in Article 5.11.2.

For Bent of Tension Reinforcements Across the Web to the Compression Face, in that case a Development Length

ld should Provide on Compression Face for Termination of Bars. Otherwise Continuity of Bended Bars would remain

Provisins of Development Length le under Article 5.11.2 :

According to Article 5.11.2.1.1 the Tension Development Length le should not be less than the product of the Basic

Development Length ldb & the Modification Factors as Specified in Articles 5.11.2.1.2 & 5.11.2.1.3. Whereas the


b)

c)

d)

e)

f)

g)

h)

i)

iv) Provisins of Article 5.11.2.2 in respect of Positive Moment Reinforcement & the Provided Reinforcement:

a) In Simple Supported Span at-least One-third or 33% of Positive Moment Reinforcement should be Extended beyond the Centerline of Support and would continue not Less than 150mm. Whereas in Continuous Spans that will be at least One-fourth or 25% of Positive Moment Reinforcement on the same Face.

b) Provided Steel (Reinforcement) Area against Maximum Positive moment for the 14,476.459

c) Provided Steel (Reinforcement) Area on Support Position of T-Girder Span 6,433.982

d) Percentage of Provided Reinforcement on Support of T-Girder in respect of 44.444 %

e) Since the Provided Reinforcement on Support is 44.444% of the Provided at Max. Positive Moment Reinforcement,

v)

a) Provided bar Diameter for Main T-Girder of Simple Supported Bridge 32 mm

b) 4

c) 50 mm

e) 50 mm

Basic Tension Development Length in mm for Different Size of Deformed Bars & Wires are;

For bar No-36 & below, ldb = 0.02Abfy/Öf/c ³ 0.06dbfy;

For bar No-43 & below, ldb = 25fy/Öf/c;

For bar No-57, ldb = 34fy/Öf/c;

For Deformed Wire, ldb = 0.36dbfy/Öf/c. Where

Ab is Cross-sectional Area of Deform Ber/Wire in mm2.

fy is Yield Strength of deform Ber/Wire in MPa.

Abfy/Öf/c is the Specified Compressive Strength of Concrete in MPa.

db is Diameter of Bar/Wire in mm.

As-pro-L/2 mm2

T-Girder Span = As-pro-L/2

As-pro-Supp mm2

both on Bottom & Top Surface = As-pro-Supp

%As-pro-Supp

the Maximum Positive Moment Reinforcement = 100*As-pro-Supp/As-pro-L/2

thus According Article 5.11.2.2 in respect of Positive Moment Reinforcement the Design is OK.

Computed value of Development Length le & Basic Development Length ldb for T-Girder Beam :

DBar

Maximum Number of Main Bars on Each Layer = NBar NBar

Clear Cover at Bottom of Main Girder = C-Cov-Bot C-Cov-Bot

Clear Cover at Top of Main Girder = C-Cov-Top C-Cov-Top


f) 38 mm

g) Provided bar Diameter for Transverse/Shear Reinforcements (Stirrups) 12 mm

h) 40.667 mm

i) 804.248

j) 1,906.000 mm

k) 1.400

l) 2,014.754 mm

m) 787.200 mm

n) 2,014.754 mm

o) Let Provide the Development Length for 32 No. Main Bars = 2025mm 2,025.000 mm

p) Since the Provided Development Length for Main Reinforcement is greater than the required Development Length,

5 Design of Shear Reinforcement & Checking of RCC Girder against Shearing Forces at Different Locations :

i)

a)

Table-2. Sum. of Max. Shear Forces Against All Applied Loads (DL & LL) on Interior Girder.


480.324 466.543 368.236 234.833 122.745 -37.401 -10.658

132.370 128.301 99.278 66.185 33.093 -7.895 0.000

248.072 239.353 177.156 106.240 35.324 -47.468 -35.592


b) Shearing Forces at Sopport Position of Girder 860.766 kN

c) 834.197 kN

Clear Cover on both Sides of Main Girder = C-Cov-Side C-Cov-Side

DStir.

Lateral Spacing of Main Bars = (bWeb-2*C-Cov-Side.-2*DStri-NBar*DBar)/(NBar-1) sLateral

Cross-sectional Area of Provided Main Deform Ber = pDBar2/4 Af-32 mm2

Depth of Concrete below the Top Horizontal Bar = hGir- C-Cov-Top.-DStri-DBar hTop-Bar

Modification Factor for Basic Development Length ldb (Since Depth below the MFactort

Top Horizontal Bars is greater than 300mm, thus Factor is for Increase of ldb).

Calculated Basic Development Length ldb for 32 No. Main Longitudinal Bar of ldb-Cal

T-Girder = MFactor*0.02Af-32fy/Öf/c

Calculated value of 0.06DBarfy; 0.06DBarfy

Computed value of Development Length, lb = 0.02Af-32fy/Öf/c ³ 0.06DBarfy; lb-req

lb-pro

thus the Design is OK in respect of Development Length for Main Reinforcement of Girder.

Calculated Factored Shearing Forces (Vu) at Different Locations due to Applied Loads (DL & LL):

Table for Max. Shear Forces at Different Locations of Interior Girder due to Factored DL, Lane-LL & Wheel-LL :



a. Wheel Live Load (WLLInt)

Vu-Supp.

Shearing Forces at Loaction 0.375m from Sopport of Girder Vu-0.375m.


d) 644.670 kN

e) 407.258 kN

f) 191.162 kN

g) (92.764) kN

h) (46.250) kN

ii) Factored Shearing Stress & Shearing Depth at Different Locations :

a)

b) - Mpa

c) 350 mm

d) Variable mm

the neutral axis between Resultants of the Tensile & Compressive Forces due Variable mm

0.72h 1,440.000 mm

Variable mm

h 2,000 mm

e) f 0.90

f)

Location Length of Width of Depth of Effective Calculated Calculated Calculated Calculated

of Section Segment T-Girder T-Girder Depth of value of value of value of value of

from from the Web Section for 0.9de 0.72h Effective Shearing

Support Earlier Tensial Shear Depth Stress

Section (h)mm mm mm mm mm mm mm

At Support 0.000 350.000 2000.000 1902.000 1711.800 1440.000 1711.800 1.596

375.000 350.000 2000.000 1902.000 1711.800 1440.000 1711.800 1.547

3050.000 350.000 2000.000 1882.800 1694.520 1440.000 1694.520 1.208

3050.000 350.000 2000.000 1851.714 1666.543 1440.000 1666.543 0.776

3050.000 350.000 2000.000 1794.000 1614.600 1440.000 1614.600 0.376

Shearing Forces at Loaction L/8 from Sopport of Girder Vu-L/8.

Shearing Forces at Loaction L/4 from Sopport of Girder Vu-L/4.

Shearing Forces at Loaction 3L/8 from Sopport of Girder Vu-3L/4.

Shearing Forces at Absolute Max. Moment Loaction (c.g. Position) of Girder Vu-c.g.

Shearing Forces at Loaction L/2 (Middle of Span) from Sopport of Girder Vu-L/2.

The Shearing Steress on Concrete due to Applied Shear Force. vu = (Vu - fVp)/fbvdv, = Vu/fbvdv ; Since Vp = 0;(AASSHTO-LRFD-5.8.2.9).Here,

Vp is Component of Prestressing applied Forces. Vp

For Nonprestressing RCC Structural Component the value of Vp = 0

bv is Width of T-Girder Web = bWeb. mm, bv

dv is Effective Shear Depth taken as the distance measured perpendicular to dv

0.9de

to Flexural having the greater value of either of 0.9de or 0.72h in mm. Here,

i) de is Effective Depth of Tensile Reinforcement for the Section in mm de

ii) h is Depth of T-Girder, hGir = 2000 mm,

f is Resistance Factor for Shear = 0.90 (AASHTO-LRFD-5.5.4.2).

Table for Computation of values of vu, de, bv, 0.9de , 0.72h & dv at different Section of Girder.

Table-3 Values of v u, d e, b v, 0.9d e , 0.72h & d v at different Section of Girder.

(bv) (de) (dv) (vu)N/mm2

L0.375m

L1/8

L1/4

L3/8


3050.000 350.000 2000.000 1808.222 1627.400 1440.000 1627.400 -0.090

g) Shearing Stress due to Applied Factored Shearing Forces at Different Sections of Girder :

i) Shearing Stress due to Applied Shearing Force at Support Position of Girder 1.596

ii) Shearing Stress due to Applied Shearing Force at a distance 0.375m from 1.547

1.208

0.776

0.376

(0.090)

iii) Factored Shearing Resistance for a Section under provision of AASHTO-LRFD-5.8.2.1-(Equ-5.8.2.1-2) :

a)

b) N

c) N

d) 0.90

iv)to Factored Shear Forces under Provisions of AASHTO-LRFD-5.8.2.4 :

a)

b)

c)

L1/2

vu-Supp. N/mm2

= Vu-Supp/fbvdv N/mm2

vu-0.375m. N/mm2

Support = Vu-0.375/fbvdv N/mm2

iii) Shearing Stress due to Applied Shearing Force at a distance L/8 from vu-L/8. N/mm2

Support = Vu-L/8/fbvdv N/mm2

iv) Shearing Stress due to Applied Shearing Force at a distance L/4 from vu-L/4. N/mm2

Support = Vu-L/4/fbvdv N/mm2

v) Shearing Stress due to Applied Shearing Force at a distance 3L/8 from vC-3L/8. N/mm2

Support = Vu-3L/8/fbvdv N/mm2

vi) Shearing Stress due to Applied Shearing Force at a distance L/2 from vu-L/2. N/mm2

Support = Vu-L/2/fbvdv N/mm2 . (The (-) ve. velue is not applicable.)

The Factored Shear Resitance at any Section of Component is Expressed by the Equation-5.8.2.1-2. Having the

value, Vr = fVn in which;

Vr is the Factored Shear Resitance at a Section in N Vr.

Vn is Nominal Shear Resitance in N according to AASHTO-LRFD-5.8.3.3. Vn.

f is Resistance Factor according to AASHTO-LRFD-5.5.4.2. f

Computation of values of q & b to Calculate the Nominal Shear Resistance (Vn) at Different Locations due

The Nominal Shear Resistance Vn at any Section of Girder is the Lesser value Computed by the Equations

i) Vn = Vc + Vs + Vp (Equ. 5.8.3.3-1) &

ii) Vn = 0.25f/cbvdv + Vp, (Equ. 5.8.3.3-2) in which,

Vc is Nominal Shear Resistance of Conrete in N having value = 0.083bÖf/cbvdv, (Equ. 5.8.3.3-1);

Vs is Shear Resistance Provided by Shear Reinforcement in N having value = Avfydv(cotq + cota)sina /s (Equ. 5.8.3.3-3) in which,


d)

e)

f)

g)

a 90

h)

i) - N

v)

a)

b)Min.Shear Reinforcement).

c) Min.Shear Reinforcement);

d)

e) 350,000

f) -

g)

6,433.982

6,433.982

8,042.477

11,259.468

16,084.954

14,476.459

h) - MPa.multiplied by Locked-in differencein Strain between the Prestressing Tendons

s is Spacing of Stirrups in mm;

b a is Factor for the Diagonally Cracked Concrete to transmit Tension as per AASHTO-LRFD-5.8.3.4;

q is Angle of Inclenation of Digonal Compressive Stress in ( 0 ) as per AASHTO-LRFD-5.8.3.4;

a is Angle of Inclenation of Transverse/Shear Reinforcement to Longitudinal Bars in ( 0 ); AASHTO-LRFD-5.8.3.4.

For Vertical Transverse/Shear Reinforcement the Angle of Inclenation, a = 900 0

Av is Area of Shear Reinforcement within a distance s in mm;


For Nonprestressing RCC Structural Component, the value of Vp = 0.

Computation of Value b & q at different Locations of Girder as per AASHTO-LRFD-5.8.3.4:

Calculation of Longitudinal Strain in Web Reinforcement εs in mm/mm on the Flexural Tension side of Girder withEquations ;

ex = (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/2(EsAs + EpAps); (Equ-5.8.3.4.2-1 for the Case with at Least the

ex = (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/(EsAs + EpAps); (Equ-5.8.3.4.2-2 for the Case with Less then the

ex= (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/(2(EsAs + EpAps); (Equ-5.8.3.4.2-3 for the Cases when value of es is (-) ve.in Equ.5.8.3.4.2-1& Equ.5.8.3.4.2-2). Where,

Ac is Area of Concerte on Flexural Tention side of Girder in mm2 having value Ac mm2

Ac = bWeb* hGir,/2 mm2

Aps is Area of Prestressing Steel on Flexural Tention side of Girder in mm2. Aps mm2

For RCC Structure, the value of Aps = 0

As is Area of Non-Prestressing Steel on Flexural Tention side of Girder for the

in mm2 under Consideration having respective values of Steel Area.

i) On Support Position value of Provided Steel Area in mm2 As-Sup. mm2

ii) At Distance 0.375m from Support, Provided Steel Area in mm2 As-0.375m. mm2

iii) At Distance L/8 from Support, Provided Steel Area in mm2 As-L/8. mm2

iv) At Distance L/4 from Support, Provided Steel Area in mm2 As-L/4. mm2

v) At Distance 3L/8 from Support, Provided Steel Area in mm2 As-3L/8. mm2

vi) At Distance L/2 from Support, Provided Steel Area in mm2 As-L/2. mm2

fpo is a Parameter for Modulus of Elasticity of Prestressing Tendons which is fpo.


and Surrounding Concrete (Mpa). For the usal level of Prestressing, the value

i) - N (-) for the case of Compressive due to Prestressing.

j)

k)

I)

Location Length of Factored Factored Calculated Calculated Effective

of Section Segment Moment Shear value of value of Moment

from from the for the for the Effective value

Support Earlier Section. Section. Shear Depth

Section

mm kN-m kN mm kN kN-m

At Support 0.000 0.00 860.766 1711.800 1473.459 1473.459

375.000 278.14 834.197 1711.800 1427.978 1427.978

2675.000 3,003.20 644.670 1694.520 1092.406 3003.199

3050.000 5,131.01 407.258 1666.543 678.713 5131.011

3050.000 6,318.42 191.162 1614.600 308.650 6318.423

3050.000 6,630.45 -46.250 1627.400 -75.267 6630.449

vi)

a) 0.076

b) 0.076

c) 0.058

d) 0.037

e) 0.018

f) (0.004)

vii)

recommended = 0.7fpu for both Pretensioned & Post-tensioned Case.

For Nonprestressed RCC Structural Component, the value of fpo = 0.

Nu is Factored Axil Force in N, Value will be (+) ve for the case of Tensile & Nu

For Nonprestressing RCC Structural Component, the value of Nu = 0.

Mu is Factored (+) ve Moment quantity of the Section in N-mm but not less then Vudv.

Vu is Factored Shear Force (Only (+) ve values are applicable) for the Section in N.

Table Showing the values of Mu, Vu, Computed value of Vudv & Effective value of Vudv.

Table-4. Showing the values of Mu, Vu, Computed value of Vudv & Effective value of Mu-Eff ³ Vudv.

Vudv

MuEff

(Mu) (Vu) (dv) (Vudv) (MuEff)

L0.375m

L1/8

L1/4

L3/8

L1/2

Value of vc/f/c (Ratio of Shearing Stress & Concrete Compressive Strength) at Different Section of Girder :

Value of vc/f/c at Support Pisition of Girder = vc-Sup/f/

c vc-Supp/f/c

Value of vc/f/c at a distance 0.375m from Support = vc-0.375/f/

c vc-0.375/f/c

Value of vc/f/c at a distance L/8 from Support = vc-L/8/f/

c vc-L/8/f/c


c vc-L/4/f/c

Value of vc/f/c at a distance 3L/8 from Support = vc-3L/8/f/

c vc-3L/8/f/c


c vc-L/2/f/c

Values of esx1000 with at Least the Min. Transverse/Shear Reinforcement under AASHTO-LRFD-5.8.2.5 &


provisions of Equation 5.8.3.4.2-1 :

a)

b) 3.980

c)

d) 1.000

d) 0.969

e) 0.950

f) 0.864

g) 0.667

h) 0.688

i)

j)

Equation No. 5.8.3.4.2-4 & Table -5.8.3.4.2-2.

viii) provisions of Equation 5.8.3.4.2-2 :

a)

b) 3.980

c) 0.002

ix)

Since for RCC Girder values of Prestressing Components Nu, Vp, Asp, fpo, Ep etc. are = 0, thus equation

ex = (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/2(EsAs + EpAps) stands to

ex = (Mu/dv + 0.5Vucotq )/2EsAs

Considering the Initial value of ex = 0.001at Support Position & the Effective cotq

Moment Mu-Eff. From the Equation Cotq = (ex2EsAs - Mu/dv)/0.5Vu

Having the value of cotq = 2.241, the values of exx1000 at Different Locations are ;

At Support (Bearing Center) Position = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Suport*1000

At 0.375m from Support = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-0.375*1000

At a Distance L/8 from Support = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-L/8*1000


At a Distance 3L/8 from Support = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-3L/8*1000


Since at Absolute Max. Moment Position & at L/2 Distance from Support the Shear Forces are of (-) ve value, thus

calculations for the values of ex-x1000 these Locations are not essential.

Since in all Sections from Support to L/2 the values of es-x1000 ≤ 1.00, thus values of q & b for the Sections can be

obtain from Table-5.8.3.4.2-1 in respect of values of vc/f/c. For Sections having the es-x1000>1.00, for those Cases

values of q & b can obtain in respect of value of Crack Spacing Parameter sxe, using the Equation .No.5.8.3.4.2-2,

Values of esx1000 with Less than Min. Transverse/Shear Reinforcement under AASHTO-LRFD-5.8.2.5 & the

Since in RCC Girder the values for Prestressing Components Nu, Vp, Asp, fpo, Ep etc. are = 0, thus equation

ex= (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/(EsAs + EpAps) stands to

ex = (Mu/dv + 0.5Vucotq )/EsAs

Considering the Initial value of ex = 0.002 at Support Position & the Effective cotq

Moment Mu-Eff. From the Equation Cotq = (exEsAs - Mu/dv)/0.5Vu

Values of exx1000 at Location = (Mu/dv + 0.5Vucotq )/EsAs ex-*1000

Computation of values of Crack Spacing Parameter, sxe for Sectiona are :


a)

b) - mmReinforcements on Vertical Faces (as Shrinkage & Tempeture Reinforcement)

402.124

- mm

s 232.857 mm

-

0.003dvs<As s-Applicable

c) 20 mm

d) - mm

f)

x)

a)

Location of Section From Referece Value of Support of Table q b

a) At Support Position 5.8.3.4.2.-1 0.076 1.000 NA 36.40 2.23

5.8.3.4.2.-1 0.076 0.969 NA 36.40 2.23

c) At L/8 Distance of Support 5.8.3.4.2.-1 0.058 0.950 NA 36.40 2.23

d) At L/4 Distance of Support 5.8.3.4.2.-1 0.037 0.864 NA 36.40 2.23

e) At 3L/8 Distance of Support 5.8.3.4.2.-1 0.018 0.667 NA 33.70 2.38

f) At L/2 Distance of Support 5.8.3.4.2.-1 -0.004 0.688 NA 33.70 2.38

b)

1.356

1.356

The Crack Spacing Parameter of a Section, sxe = sx(35/(ag+16)) £ 2000mm, In which;

sx = the Lesser value of either dv or the Spacing of Longitudinal Crack Control sx

having Steel Area in a Horizontal Reinforcement Layer if As > 0.003bvs. Here;

i) As is Steel Area of 2nos Shrinkage & Tempeture Reinforcement Bars on As mm2

Opposite Vertical Faces of T-Girder in same Horizontal Layer,

= As-S&T-Hori. = 2*pDBar2/4 = 2*Af-16

ii) dv is Effective Shear Depth of Tensil Reinforcement for the Section dv

iii) s is Spacing of Longitudinal Bars as Shrinkage & Tempeture Reinforcement on Vertical Faces of T-Girder.

iv) Thus Computed value of 0.003dvs for a Section 0.003dvs mm2

v) For Computed value of 0.003bvs at Section > As, the value of sx = dv

ag is Max. Aggragate size for Concrete = 20mm ag

Thus Computed value of sxe = sx(35/(ag+16)) for Section at L/8 sxe

For value of ex-*1000 > 1.00 & the value of sxe at a Sections should Computed from the Respective Tablel.

Computation of Values of q & b from AASHTO-LRFD'sTable -5.8.3.4.2-1. & Table -5.8.3.4.2-2. against the

Respective Calculated values of esx1000, Ratio vc/f/c & sxe.:

Table for Values of q & b at Different Location of Girder.

vc/f/c exx1000 sxe

b) At L0.375mDistance of Support

Value of Cotq at different Locaion of Girder :

i) At Support (Bearing Center) Position value of Cotq Cotq

ii) At a Distance 0.375m from Support value of Cotq Cotq


1.356

1.356

1.499

1.499

xi)

a)

b) 18,452.293 kN AASHTO-LRFD-5.8.3.3-(Equ. 5.8.3.3-1); 18452.293*10^3 N

xii) Regions Requiring Transverse or Shear/Web Reinforcements under AASHTO-LRFD-5.8.2.4 :

a) The Transverse or Shear Reinforcements are required for those Sections where the Factored Shearing Force due to

b)

c)Equation-5.8.3.3-4.

d) - N

e)

f) f 0.90

g)

h)

Location from Girder's Values of Relation Equation

Bearing Center. b Between Satisfied/

N N N Not Satisfied

a) At Support (Bearing Center) Position 2.230 860765.82 48903800.52 22,006,710 Vu<0.5Vc Not Satisfied

a) At a Distance 0.375m from Support 2.230 834196.637 48903800.52 22006710.23 Vu<0.5Vc Not Satisfied

2.230 644669.804 48410134.39 21784560.48 Vu<0.5Vc Not Satisfied



2.380 -92764.25 49619912.6 22328960.7 Vu<0.5Vc Not Satisfied

iii) At a DistanceL/8 from Support value of Cotq Cotq

iv) At a Distance L/4 from Support value of Cotq Cotq

v) At a Distance 3L/8 from Support value of Cotq Cotq

vi) At a Distance L/2m from Support value of Cotq Cotq

Computation of Value for Nominal Shearing Strength of Concrete (Vc) using the Values of q & b :

Since the Location of L/8 is next to the Critical Section at a Distance 0.375m from Support, thus Values of q & b of

Section L/8 are the Governing Values for Computation of Nominal Shearing Strength of Concrete (Vc) for the Girder.

Nominal Shear Resistance of Conrete of Girder Vc = 0.083bÖf/cbvdv, Vc

the Applied Loads (DL & LL), Vu > 0.5f (Vc + Vp); AASHTO-LRFD-5.8.2.4; Equ-5.8.2.4-1; Here,

Vu is Factored Shearing Force due to the Applied Loads for the Selected Section in N,

Vc is Nominal Shear Resistance for the Section having value = 0.083bÖf/cbvdv according to AASHTO-LRFD-5.8.3.3.



b a is Factor for the Diagonally Cracked Concrete to transmit Tension according to AASHTO-LRFD-5.8.3.4;

f is Resistance Factor according to AASHTO-LRFD-5.5.4.2. having value = 90

Thus for Nonprestressing Structure the Eqution-5.8.2.4-1 Stands to Vu > 0.5Vc

Table showing the values of b, Vu, Vc, 0.5Vc & Relation between Vu & 0.5Vc at Different Location of Girder.

Vu Vc 0.5fVc

Vu & Vc

b) At a DistanceL/8 from Support

c) At a Distance L/4 from Support

d) At a Distance 3L/8 from Support

e) At a Distance L/2m from Support


i)

xiii) Checking of Required Max. Spacing for Transvers/Shear Reinforcement due to Applied Shearing Stress on Girder under provision of AASHTO-LRFD-5.8.2.7 :

a)

b)

c) 2.625 due to Applied Shearing Stress at Defferent Section of Girder.

d)

Table- ; Showing Spacing of Transverse/Shear Reinforcements in respect of Max. Spacing & Status :

Segment. Value of Value of Value of Relation Value of Value of Relation Relation Formula

Location between between between Status

(Between for the for the for the Whether

2-Section) Section Maximum Section Section Satisfy or

mm mm mm Not Satisfy

Section at 1711.80 1.596 2.625 vu<0.1.25f/c 1369.440 684.720 0.8dv>600 Not Satisfy

0.4dv>300 Not Satisfy

Section at 1,711.80 1.547 2.625 vu<0.1.25f/c 1369.440 684.720 0.8dv>600 Not Satisfy

L/8 0.4dv>300 Not Satisfy




3L/4 0.4dv>300 Not Satisfy



e)

xix) Chacking for Transverse/Shear/Web Reinforcements as Deep Beam Component (AASHTO-LRFD-5.8.1.1) :

a)Support is Considered as a Deep Component. Deep Beam Components the Shear Reinforcements are being Provide

b) 860.766 kN

The Table indicates that non of the Sections of RCC Girder have satisfied the required provisions for Transverse/Shear Reinforcements under Equation Vu > 0.5f (Vc + Vp). AASHTO-LRFD-Equation-5.8.2.4-1.

Due to applied Shearing Stress, vu < 0.125f/c, the Max. Spacing of Transverse/Shear Reinforcement at a Section is

smax.-1 = 0.8dv ≤ 600mm, (AASHTO-LRFD-Equ. 5.8.2.7-1).

Due to applied Shearing Stress, vu > 0.125f/c, the Max. Spacing of Transverse/Shear Reinforcement at a Section is

smax.-2 = 0.4dv ≤ 300mm (AASHTO-LRFD-Equ. 5.8.2.7-2)

Value of 0.125f/c in respect of Max. Spacing of Transvers/Shear Reinforcement 0.125f/

c N/mm2

Table Showing values of vu, 0.8dv, 0.4dv againest the respective values of 0.125f/c :

dv vu 0.125f/c 0.8dv 0.4dv

vu & 0.12f/c 0.8dv & 0.4dv &

sMax-1<600 sMax-2<300

N/mm2 N/mm2 vu<=>0.12f/c

L0.375m

The Table indicates that non of the Sections of RCC Girder have satisfied the required provisions for Spacings of the

Transverse/Shear Reinforcement under Equations, For vu < 0.125f/c ; smax. = 0.8dv ≤ 600mm, (AASHTO-LRFD-Equ.

5.8.2.7-1). & For vu > 0.125f/c ; smax. = 0.4dv ≤ 300mm, (AASHTO-LRFD-Equ.5.8.2.7-2).

The Component in which a Load causing more than 1/2 of the Shear at a Distance closer than 2d from the Face of

according to Provisions of AASHTO-LRFD-5.6.3 (Provisions of Strut-and-Tied Model) & AASHTO-LRFD-5.13.2.3.

Factored Shear Force at Support of Girder, = Vu-Supp. kN. Vu-Supp.


c) 430.383 kN

d) 608.249 kN

e) Vu-2d>Vu-Supp.

f)

can Provide.

xx) Detaling of Requirments for Deep Beam Component to Provide Transverse/Shear/Web Reinforcements for Girder under provision of AASHTO-LRFD-5.13.2.3 :

a) To provide Transverse/Shear/Web Reinforcements at Different Section of Girder, it should Satisfy the Equation,

b) N

c) 350 mm

d) 410.000 MPa

e)

f) 300 mm

g) 300 mm

xxi) Computation of Spacing for Transverse/Shear Reinforcement at Different Section of Girder :

a) d/4 500 mm

b) Allowable Max. Spacing for Transverse/Shear Reinforcement 300 mm

c) 12 mmof Vertical Stirrups.

d) 226.195

d) Let Provide the 175mm Spacing for Transvers/Shear Reinforcement from Outer 175 mm

1/2 of Factored Shear Force at Support of Girder, = 1/2Vu-Supp. kN. 1/2Vu-Supp.

Factored Shear Force at a Distance 2d from Support, = Vu-2d. kN. Vu-2d.

Status of 1/2Vu-Supp. & Vu-2d

Since the Factored Shear Force at a Distance 2d, Vu-2d > 1/2Vu-Supp. the Max. Shear Force at Support, thus theGirder is Deep Beam Component & under Article 5.13.2.3. Of AASHTO-LRFD its Transverse/Shear Reinforcement

Ng = f fyAs ³ 0.83bvs (AASHTO-LRFD-5.13.2.3). Here,

Ng is Factored Tensial Resitance of Transverse Reinforcement (Each Pair) in N. Ng.

bv is Width of Girder Web in mm. bv.

fy is Yield Strength of Reinforceing Steel as Transverse/Shear Reinforcement fy.

As is Steel Area of Transverse/Shear Reinforcement in mm2 having Spacing s . AS mm2

s is Spacing of Transverse /Shear Reinforcement in mm. The value of s should s-Max

not excide eithe of d/4 or 300mm

The Vertical Spacing, sVetrt. for Crack Control Longitudinal Reinforcement on both sVert.-Max

Vertical Faces of Girder should not Excide either d/3 or 300 mm.

Value of d/4 for the Section under cosideration having d = hGir. (Girder Depth).

s-Max

Let Provide 2-Leged 12f bars as Transverse/Shear Reinforcement in the form DStir.

X-Sectional Area of 2-Leged 12f Bars as Transverse/Shear Reinforcement; Av mm2

= 2*pDStir.2/4 mm2

s-0.375m.

Face of Girder up to Support Face at Section at L0.375m Distance.


e) Let Provide the 175mm Spacing for Transvers/Shear Reinforcement in between 175 mm

i) Let Provide the 200mm Spacing for Transvers/Shear Reinforcement in between 200 mm

j) Let Provide the 200mm Spacing for Transvers/Shear Reinforcement in between 200 mm

j) Let Provide the 200mm Spacing for Transvers/Shear Reinforcement in between 200 mm

xxi) Checking for Requirements of Minimum Transverse Reinforcement for Different Section of Girder:

a)

b) Table for Minimum Transverse/Shear Reinforcement at Different Section of Girder.

Location from Bearing Center of Length of s' Web Bar Status of

Girder. Segment Width Spacing Required Provided Provided &

mm mm mm

a)Between outer Face to Support Face 300.000 350.000 175.000 56.821 226.195 Av-pro> Av

3050.000 350.000 175.000 56.821 226.195 Av-pro> Av

3050.000 350.000 200.000 64.938 226.195 Av-pro> Av

3050.000 350.000 200.000 64.938 226.195 Av-pro> Av

3050.000 350.000 200.000 64.938 226.195 Av-pro> Av

xxii)

a)

b) 1,230.434 kN 1230.434*10^3 N

c) 1,230.434 kN 1230.434*10^3 N

d) 1,230.434 kN 1230.434*10^3 N

e) 1,065.762 kN 1065.762*10^3 N

s-L/8

Support Face Section at L0.375m & Section at L/8 Distance

s-L/4

Section at L/8 & Section at L/4 Distance

s-3L/8

Section at L/4 & Section at 3L/4 Distance

sL/2

Section at 3L/4 & Section at L/2 Distance

Minimum Transverse/Shear Reinforcement at a Section, Av ³ 0.083Öf/c(bvs/fy), AASHTO-LRFD-5.8.2.5.

Where, s is Length of Girder Segment under consideration for Transverse/Shear Reinforcement.

bv' Web Min. Av Av-pro

mm2 mm2 Required Av

b)From Support Face to L/8 of Girder

c) From L/8 to L/4 of Girder Length

d) From L/4 to 3L/4 of Girder Length

e) From 3L/4 to L/2 of Girder Length

Computation of Values of Vs,the Shear Resistance against Provided Shear Reinforcement & Spacings at

Different Sections according to Vs = Avfydv(cotq + cota)sina /s (AASHTO-LRFD-Equ. 5.8.3.3-3) :

With Vertical Shear Reinforcement the value of a = 900 & the Equation Vs = Avfydv(cotq + cota)sina /s stands to

Vs = Avfydvcotq /s,

At Support Position (Bearing Center) of Girder Vs= Avfydv-Supp.Cotq/sL0.375m VS-Supp.

At distance 0.375m from the Support of Girder Vs=Avfydv-L0.375mCotq/sL0.375m VS-L0.375m

At a distance L/8 from the Support of Girder Vs = Avfydv-L/8Cotq/sL/8 VS-L/8

At a distance L/4 from the Support of Girder Vs = Avfydv-L/4Cotq/sL/4 VS-L/4


f) 1,158.726 kN 1158.726*10^3 N

g) 1,122.611 kN 1122.611*10^3 N

xxii)

a)

b)

c) 50,134.235 kN 50134.235*10^3 N

d) 50,134.235 kN 50134.235*10^3 N

e) 49,640.569 kN 49640.569*10^3 N

f) 48,676.627 kN 48676.627*10^3 N

g) 50,388.363 kN 50778.639*10^3 N

h) 50,742.524 kN 50742.524*10^3 N

xxiii)

a)

b)

c) 3,145.433 kN 3145.433*10^3 N

d) 3,145.433 kN 3145.433*10^3 N

e) 3,145.433 kN

At a distance 3L/8 from the Support of Girder Vs = Avfydv-3L/8Cotq/s3L/8 VS-3L/8

At a distance L/2 from the Support of Girder Vs = Avfydv-L/8&L/2Cotq/s 3L/8&L/2 VS-L/2

Computation of values for Nominal Shear Resistance (Vn) at Different Section of Girder under Provisions

of AASHTO-LRFD-5.8.3.3 against Equation Vn = Vc + Vs + Vp (Equ. 5.8.3.3-2) :

The Nominal Shear Resistanceat any Section of Girder is Vn = Vc + Vs + Vp (Equ. 5.8.3.3-1)

For RCC Girder the value of Vp = 0, thus Equation stands to Vn-1 = Vc + Vs

At Support Position (Bearing Center) of Girder Vn= Vc-Supp + Vs-Supp. Vn-Supp.-1

At distance 0.375m from the Support of Girder Vn = Vc-L0.375m + Vs-L0.375m Vn-L0.375m-1

At distance L/8 from the Support of Girder Vn = Vc-L/8 + Vs-L/8 Vn-L/8-1


At distance 3L/8from the Support of Girder Vn = Vc-3L/8 + Vs-3L/8 Vn-3L/8-1



of AASHTO-LRFD-5.8.3.3 against Equation Vn = 0.25f/cbvdv + Vp (Equ. 5.8.3.3-2) :

According to Equ. 5.8.3.3-1 the Nominal Shear Resistanceat any Section of Girder is Vn = 0.25f/cbvdv + Vp

For RCC Girder the value of Vp = 0, thus Equation stands to Vn = 0.25f/cbvdv

At Support Position (Bearing Center) of Girder Vn= 0.25f/c-bv-Suppdv-Supp. Vn-Sup.

At distance 0.375m from the Support of Girder Vn= 0.25f/c-bv-L0.375mdv-L0.375m. Vn-L0.375m-2

At distance L/8 m from the Support of Girder Vn= 0.25f/c-bv-L/8dv-L/8 Vn-L/8-2


3145.433*10^3 N

f) 3,113.681 kN 3113.681*10^6 N

g) 3,062.273 kN 3062.273*10^3 N

h) 2,966.828 kN 2966.828*10^3 N

xxix)Computed accordting Provisions of AASHTO-LRFD-5.8.3.3 & AASHTO-LRFD-5.8.2.1 :

a)below ;

b)

c)

d)

e)

f) N

g) 0.90

h)

i)

Section Calculated Relation Accepted Factored Relation Status

Location Factored As per As per Equ. between Value of Shear between If Vr > Vu, the

from Center Shear Equation. 5.8.3.3-2 Values of Resitance Values of Structure is

of Bearing. 5.8.3.3-1 Safe otherwise

kN kN kN kN kN No Safe.

At Support 860.766 50134.235 3145.433 Vn-1> Vn-2 3145.433 2,830.889 Vr> Vu Structure Safe

834.197 50134.235 3145.433 Vn-1> Vn-2 3145.433 2,830.889 Vr> Vu Structure Safe

At L/8 644.670 49640.569 3145.433 Vn-1> Vn-2 3145.433 2,830.889 Vr> Vu Structure Safe

At L/4 407.258 48676.627 3113.681 Vn-1> Vn-2 3113.681 2,802.312 Vr> Vu Structure Safe

At 3L/4 191.162 50388.363 3062.273 Vn-1> Vn-2 3062.273 2,756.045 Vr> Vu Structure Safe

At L/2 -46.250 50742.524 2966.828 Vn-1> Vn-2 2966.828 2,670.145 Vr> Vu Structure Safe


At distance 3L/4 m from the Support of Girder Vn= 0.25f/c-bv-3L/4dv-3L/4 Vn-3L/8-2


Accepted Nominal Shear Resistance-Vn & Factored Shearing Resistance-Vr at Different Section of Girder

Nominal Shear Resistance, Vn at any Section of Component is the Lesser value of of the Equqtions as mentioned

Vn = Vc + Vs + Vp (Equ. 5.8.3.3-1; AASHTO-LRFD-5.8.3.3)

Vn = 0.25f/cbvdv + Vp (Equ. 5.8.3.3-2; AASHTO-LRFD-5.8.3.3) :

For RCC Girder, Vp = 0.





The Acceptable Nominal Shear Resistance, Vn, Respective Factored Shear Resitance-Vr at Different Section of

T-Girder, the Staus between the vaues of Vn Computed under Equ. 5.8.3.3-1 & Equ. 5.8.3.3-2 are shown in Table below :.

Table-; Accepted Nominal Shear Resistance-Vn & Factored Shear Resitance-Vr at Different Section :

Vn-1 Vn-2

Vn

Force-Vu Vn-1 & Vn-2 Vr Vr& VU

Vn-1 > Vn-2 Vr> VU

At L0.375m


xxx)of Girder, Computed under Provisions of AASHTO-LRFD-5.8.3.3 :

a)are shown in the Table blow :

Location of Section from Bearing Calculated Computed Status Center of Girder. value of value of between

Values of

kN kNa) At Support (Bearing Point) Position 860.766 3145.433 Vu< Vn

834.197 3145.433 Vu< Vn

644.670 3145.433 Vu< Vn

407.258 3113.681 Vu< Vn

191.162 3062.273 Vu< Vn

-92.764 2966.828 Vu< Vn

b)is Safe in respect of Applied Shearing Forces caused by Dead Load & Live Loads to the Bridge Structure.

xxxi)

a)Support.

b) 1,711.800 mm

c) 2.087 m

e) 3,145.433 kN

3145.433*10^3 N

f) Since the T-Girders are being provided equal Web Width through the Length, 350 mm

g) 350 mm

h)

Relation between Factored Shearing Force (Vu) & the Nominal Shear Resistance (Vn) at Different Section

The Relation between Factored Shearing Force, Vu & the Nominal Shear Resistance Vn at Different Section of Girder

Table-; Showing between Factored Shearing Force, Vu & the Nominal Shear Resistance Vn :

Vu Vn

Vu & Vn

b) At Distance L0.375m from Support.c) At Distance L/8 from Support.d) At Distance L/4 from Support.e) At Distance 3L/4 from Support.f) At Distance L/2from Support.

Since the Factored Shearing Forces Vu < Vn < Vr, the Computed Nominal Shear Resistance, thus the Girder

Checking of T-Girder Web Width (bWeb) in respect of Nominal Shearing Resistance (Vn) at Critical Section :

According to AASHTO-LRFD-5.8.3.2 the Critical Section for Shearing Forces Prevails at a Distance dv from Face of

Since In-between Support & the Section at L/4 the value of dv is same, thus on dv

Critial Section value of dv = 1697.400mm

Distance of Critical Section from Support/Bearing Point, LCrit = (0.375 + dv ) LCrit.

Since the Nominal Shearing Resistance Vn for Girder In-between Support & the Vn-Crit.

Section at L/4 have same values, thus at Critical Section Vn will have same value.

bv-Crit.

thus Web Width for Critical Section, bWeb = 350 mm which is the Effective Web

Width bv for the Section.

According to Equ. 5.8.3.3-2, at any Section of Girder the Effective Web Width, bv-Crit-Cal.

bv = (Vn - Vp)/0.25*f/cdv. In a RCC Girder the value of Vp = 0. and the Equation

Stands to, bv = Vn/0.25*f/cdv. from which value of bv-Crit can Calculate.

Since the Calculated value of Effective Web Width for Critical Section bv-Crit = bWeb the Provided Web Width, thus


xxxii) Checking in respect of Longitudinal Reinforcements (Tensile Reinforcements) Provided for Girder under Provision of AASHTO-LRFD-5.8.3.5 :

a) At each Section the Tensial Capacity of Longitudinal Reinforcement on Flexural Tension side of the Member shall be

Where;

b) Variable

c) 410.000 MPa

d) -

e) - MPa

f) Variable N-mm

g) - N(-) for the case of Compressive due to Prestressing.

h) Variable mm

i) Variable N

j) Variable N

k) - N

l) q VariableAASHTO-LRFD-5.8.3.4;

m) 0.90 AASHTO-LRFD-5.5.4.2.

n) 0.90 AASHTO-LRFD-5.5.4.2,

o) 0.80

Design of Critical Section is OK.

proportationed to Satisfy Equation, Asfy + Apsfps ³ Mu/dvff + 0.5Nu/fc + (Vu/fv - 0.5Vs -Vp)Cotq. (Equ-5.8.3.5-1).

As is Area of Nonprestressing Steel on Flexural Tention side of Girder in mm2 As mm2

fy is Yield Strength of Nonprestressing Reinforceing Steel in MPa. fy


For Nonprestressing RCC Structure, the value of Aps = 0

fps is Yield Strength of Prestressing Steel in MPa. fps

For Nonprestressing RCC Structure, the value of fps = 0

Mu is Factored Moment of the Section due to Dead & Live Loads Loads on Mu

Structure. in N-mm but not less then Vudv.



dv is Effective Shear Depth of Tensil Reinforcement for the Section in mm. dv

Vu is Factored Shear Force for the Section in N. Vu

Vs is Shear Resistance Provided by Shear Reinforcement in N for the Section. Vs



q is Angle of Inclenation of Digonal Compressive Stress in ( 0 ) according to O

ff is Resistance Factor for Flexural Tension of Reinforced Concrete according to ff

fv is Resistance Factor for Shearing Force of Reinforced Concrete according to fv

fc is Resistance Factor for Compression due to Prestressing according to fc


AASHTO-LRFD-5.5.4.2.

p)

q) Table- Showing Evalution of Equation-5.8.3.5-1 at Different Section of Girder & Status of Results.

Location of Section Calculted Calculted Status

from Bearing Center Provided Factored Factored Shearing Value of R/H Part of the

of Girder. Tensile Moment Shearing Resistance of the EquationSteel Area Force of Stirrups (L/H Part) Equation Satisfied

kN-mm kN kN kN kN Not Satisfieda)At Support 6433.982 1473.459 860.766 1230.4342612 2637.933 1,237.467 Satisfied

6433.982 1427.978 834.197 1230.4342612 2637.933 1,173.513 Satisfied

8042.477 3003.199 644.670 1230.4342612 3297.416 1,732.176 Satisfied

11259.468 5131.011 407.258 1065.7617895 4616.382 2,616.188 Satisfied

16084.954 6318.423 191.162 1158.7263044 6594.831 2,861.967 Satisfied

14476.459 6630.449 -92.764 1122.6110886 5935.348 2,699.711 Satisfied

j) Since at all Section the requirments of Equation are being Satisfied, thus provision of Transverse/Shear Reinforcements for Girder is OK.

xxxiii) Checking for Factored Tensile Resistance of Transverse/Shear Reinforcements Provided for Girder:

a) The provided Transverse/Shear/Web Reinforcements at Different Section of Girder under the Provision of Deep Beam

b) Variable N

c) 350 mm

d) 410.000 Mpa

e) 226.195

f) s Variable mm

g) The Girders are being provided with Transverse/Shear Reinforcements in the form of Vertical Stirrups against Shear Forces caused by the Dead Load & Live Loads applied to the Bridge Structure. Due to Vertical Positioning of those Transverse/Shear Reinforcements they will also under Tensile Stress caused by the same Shearing Forces whose actions are in some extent in Vertical Direction. On these back drop mentioned Shearing Forces at different Sectionof Girder may be considered as Tensile Forces for Vertical Stirrups. Thus the Factored Shearing Forces at Each

Since for Nonprestressing RCC Structural Components the Items Aps, fps, Nu & Vp have Values = 0, thus mentioned

Equ-5.8.3.5-1. Stands to, Asfy ³ Mu/dvff + (Vu/fv - 0.5Vs)Cotq.

As. Mu Vu Vs

Asfy

mm2

b) At L0.375m.c) At L/8.d) At L/4.e) At 3L/4.f) At L/2.

should Satisfy the Equation, Ng = f fyAs ³ 0.83bvs (AASHTO-LRFD-5.13.2.3; Equ-5.13.2.3-1). Here,




As is Steel Area of Transverse/Shear Reinforcement in mm2 having the Spacing AS mm2

s mm. Here As = Av, the X-Sectional Area of 2-Leged the 12f Dia. Vertical Stirrups.

s is Spacing of Transverse /Shear Reinforcement in mm. The value s should not Excide eithe d/4 or 300mm

Section, Vu = Ng, the Factored Tensile Resistance Carried by the Pair of Transverse Reinforcement.


h) The Checking for Factored Tensile Resistance of Transverse/Shear Reinforcements at Different Section of Girder areshown in the under mention Table:

i) Table- Showing Minimum Transverse/Shear Reinforcement at Different Section of Girder.

Location of Section from Bearing s'-Spacing Calculted Calculted Status of

Center of Girder. of Shear Provided Value of Value of Equation

Bars Steel Area Satisfiedmm Not Satisfied

a) At Support (Bearing Point) Position 175.000 226.195 83465.834 50837.50 Satisfied

175.000 226.195 83465.834 50837.50 Satisfied175.000 226.195 83465.834 50837.50 Satisfied200.000 226.195 83465.834 58100.00 Satisfied200.000 226.195 83465.834 58100.00 Satisfied200.000 226.195 83465.834 58100.00 Satisfied

j)thus Provision of Transverse/Shear Reinforcements for Girder is OK.

xxxiv)Provision of AASHTO-LRFD-5.8.3.2 :

a)Support & the Area In-between this Section & Support Face should be Designed f

b)

c) 1,711.800 mm

d) 2.087 m

e) 4,515.613 kN-m

f)

g) The provided Steel Area as Tension Reinforcemen on bottom surface of Critical 6,433.982

Section. The Critical Section is being also provised same Steel Area on its Top Surface.

As

ffyAs 0.83bvsmm2

b) At Distance L0.375m from Support.c) At Distance L/8 from Support.d) At Distance L/4 from Support.e) At Distance 3L/4 from Support.f) At Distance L/2from Support.

Since in all Section the requirments of Equation Equ-5.13.2.3-1, Ng = ffyAs > 0.83bvs are being Satisfyed,

Checking the Critical Section Near Support in respect of Mn the Nominal Flexural Resistance under the

According to AASHTO-LRFD-5.8.3.2 the Critical Section for Shearing Forces Prevails at a Distance dv from Face of

According to AASHTO-LRFD-5.8.3.2 at Critical Section will have Nominal Flexural Resistance Mn= dv*(Apsfps + Asfy) in N-mm both for Top & Bottom Tension Reinforcement Bars.

Since the position of Critical Section at a Distance dv from Support is related to dv-Crit

Distance of the Sections fron Support to 0.375m & L/8. The Calculated Value

of dv for Section at Support, at 0.375m, L/8 & L/4 are same. Thus for Critical

Section the value of dv within mentioned Sections are Applicable.

Distance of Critical Section from Support/Bearing Point, LCrit = (0.375+ dv ) LCrit.

For Nonprestrssed RCC Structure value of Aps = 0 & also fps = 0; thus for the Mn-Crit-1.

Critical Section Equation Mn = dv(Apsfps + Asfy) stands to Mn= dvAsfy.


Section the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2) in which,

As-Crit. mm2

Section have same value that for the Sections In-between Support Point & L/8.


h) 1,882.800 mm

i) 21.266 mm

j) 4,491.197 kN-mSteel Area against Factored Max. Moments at its Critical Section will have the 4491.197*10^6 N-mm

k)Calculated value of Nominal Resistance based on provided Effective for Tensial Reinforcement for Critical Section thus

xxxv) Checking the Critical Section Near Support in respect of Provided Shear & Tensial Reinforcements for the Section according to Provisions of AASHTO-LRFD-5.8.3.2 :

a) The Shearing Resistance for the provided Shearing Reinforcement In-between Support Face & Critical Section should

b) The Calculated Shearing Forces at Critical Section due Factored Loads 673.904 kN 673.904*10^3 N

c) 1,979.602 kN-m 1979.602*10^3 N-mm

d) 1,153.589 kN-m

e) 1,979.602 kN-m 1979.602*10^3 N-mm

f) 1.250

g) 0.060

h) 0.001

i) q 24.300

j) 2.215

k) s 175.000 mm

The Effective Depth for Tension Reinforcement of Critical Section, de = ds ds-Crit.

is same that for the Sections In-between Support Point & L/8.

Accordingly the Depth of Compressin Block 'a' for Critical Section have same aCrit.

value that for the Sections In-between Support Point & L/8.

For Simple Supported & Single Reinforced T-Girder Beam the with Provided Mn-Crit-2.


Since the Calculated value of Nominal Resistance based on provided Effective Shear Depth, Mn-Crit.-1 < Mn-Crit-2 the

the Design of Critical Section is OK.

have the value Vs = AvfydvCotq/s in N/mm2.

VU-Crit.

The Calculated Moment at Critical Section due Factored Loads (DL & LL). MU-Crit.

Calculated value of VU*dv for Critical Section VU*dv

The Effcetive Moment for Critical Section is Greater One of MU-Crit & VU*dv MEff.

The Shearing Stress at Critical Section vu = VU/fbvdv N/mm2 vu N/mm2

Value of vu/f/c for Critical Section vu/f/

c

Based on the Initial Longitudinal Straion of Tensial Reinforcement ex = 0.001, ex*1000

Effcetive Moment for Critical Section MEff,Effective Depth dv & value of Cotqthe Computed value of ex*1000 = (MU/dv + 0.5VUcot q )/EsAs

Based on values of ex*1000 & vu/f/c from respective Table LRFD-5.8.3.4.2-1 O

Value of Cotq for Critical Section Cotq

Provided Spacing of 2-Legged 12f Vertical Stirrups between Support & Critical


Section.

l) 226.195

m) 2,009.124 kN2009.124*10^3 N

n) LHP 2,637.933 kNRHP 474.310 kN

whether those Satisfy the Provision of Equation or Not. LHP>RHP Satisfy

o)Shear Reinforcements as well as by the Tensial Reinforcements provided In-between Support Face & the Critical

6 Design of Reinforcement for Torsion & Checking against Forces due to Torsion at Different Locations :

i) Design Phenomenon for Tensional Effect upon Girder & Checking in these respect :

a) The Bridge Girders are being considered as T-Beam having Dead Loads due to Self Weight & Superimposed Loads from Deck Slab, W/C, Sidewalk, Curb/Wheel Guard, Railing, Rail Posts etc. whereas Live Loads from Wheel Loads of Truck, Lane Loads, Pedestrian etc. Thus the Torsional Forces at Section is the Total Shearing Forces acting on that Section.

b)Tensional Forces due to Eccentric Loading of both Dead Loads & Live Loads. Dead Loads of Structure & Live LaneLoads have almost equilibrium actions. Whereas Truck Live Loads have a significant effect in these respect. Wheel

Line of Girder of each side of Interior Girders. Thus Location or Eccentricity of Torsional Force action will be consider

c) 860.765820 kN

860765.820 N

d) 500 mm

d) 430.383 kN-m430.383*10^6 N-mm

ii)

a)

Steel Area of 2-Legged 12f Vertical Stirrups between Support & Critical Section Av mm2

Computed value of Vs = AvfydvCotq/s for the Critical Section Vs-Crit.

Computation of values of Equation Asfy ³ Mu/dvff + (Vu/fv - 0.5Vs)Cotq. for theCritical Section as LHP (Left hand Part) & RHP (Right hand Part) in respect of

Since the Provisions of Equation Asfy ³ Mu/dvff + (Vu/fv - 0.5Vs)Cotq. are being Satisfied by the arramgement of

Section, thus the Flexural Design of Critical Section is OK.

Bridge Interior T-Girders has a Flange Width, b = 2.000m, & Web bWeb = 0.450m. Girders are being effect by the

Position of a moving Truck may change over Bridge Deck & will have effect up to a distance of b/2 from the Center

at a distance b/4 from Girder Center Line.

Based on the assumption-a the Factored Torsional Forces (DL + LL) will be FTor.

the Factored Shearing Forces at a Section. Thus FTor. = FShear

Based on the assumption-b of Torsional Forces (DL & LL) action Positions, the eTor.

Eccentricity eTor. = b/4 from Center of Girder.

Accordingly Factored Torsional Moment at any Section will be Tu = FTor.*eTor. Tu.

Computation of Factored Torsional Forces (TForces) &Torsional Moments (Tu) under assumed Provisions :

According to assumption the Factored Forces for Torsion, FTor. due to Dead Load & Live Loads at Different Section

of Girder are the respective Factored Shearing Forces FShear. Whereas the Torsional Moments, Tu are due to the

Eccentric action of Torsional Forces FTor. The Factored Shear Forces FShear, Factored Torsional Forces, Fu & the

respective Torsional Moments, Tu at different Sections of Girder are shown in under mentioned Table.


b) Table-2. Factored Shear Forces, Torsional Forces & Moments at Different Sections on Interior Girder.

Locations from Support On Support 0.375m L/8 L/4 3L/8 L/2

248.07 239.353 177.156 106.240 35.324 -47.468

860.77 834.197 644.670 407.258 191.162 -92.764

430.383 417.098 322.335 203.629 95.581 -46.382

c)gradually reduced to zero or (-) ve value towards the Center of Span.

b) Since opposite direction Torsional Forces & Moments at two Support Positions with highest magnitude causes the Torsional affect on T-Girders, thus those are considered as the Governing Torsional Forces & Moments for Design.

iii)AASHTO-LRFD-5.8.2.1 :

a)

b) 430.383 kN-m

N-mm

c) 607.707 kN-m

607.707*10^6 N-mm

d) 1030000.000

e) 8,000.000 mm

f) - Mpaof Prestress Losses either at the Centroid of Cross-section Resisting Transient Loads or at the Junction of Web & Flange when the Centroid Lies within Flange

g) f 0.90

h) 136.734 kN-m 136.734*10^6 N-mm

i)

Locations from Support On Support 0.375m L/8 L/4 3L/8 L/2 Unit of Moments kN-m kN-m kN-m kN-m kN-m kN-m

Shearing Forces, FTor. (kN)

Torsional Forces, FTor. (kN)

Torsional Monent, Tu (kN-m)

From Table - 2, it appears that, magnitude of Torsional Forces & Moments are highest at Support Positions & those

Checking of Design in respect of Factored Torsional Moment (Tu) at Support Position under Provisions of

In Normal density Concrete the Factored Torsonal Moment, Tu > 0.25fTcr (Equ.-5.8.52.1-3), in which;

Tu is Calculated Factored Torsional Moment at Support Position in N-mm. Tu

430.383*106

Tcr is Torsional Cracking Moment of Component in N-mm. having the value, Tcr

Tcr = 0.328Öf/c*(Acp

2/Pc)Ö(1 + fpc/0.328Öf/c) N-mm. where,

Acp is Total Area Enclosed by the Outside Perimeter of the Concrete Section in Acp mm2

mm2; Here for T-Girder Section Acp = b*hf +bWeb*(hGir. - hf) mm2

Pc is Length of Outside Parameter of Concrete Section in mm. Pc

Here for T-Girder Section Pc = 2*(b + hf) - bWeb + 2*(hGir - hf) + bWeb mm

fpc is Compressive Stress in Concrete Section for Prestressing after occurrence fpc

having unit of value in Mpa. For Nonprestressing RCC Structural Components the

value of fcp = 0.

f is Resistance Factor for Torsion according to AASHT-LRFD-5.5.4.2.

Calculated value of 0.25f Tcr for T-Girder 0.25fTcr

Table-2. Status of Factored Moments (T u) in respect of Torsional Cracking Moment (T cr)at Different Section.


430.383 417.098 322.335 203.629 95.581 -46.382

136.734 136.734 136.734 136.734 136.734 136.734

Tu>0.25fTcr Tu>0.25fTcr Tu>0.25fTcr Tu>0.25fTcr Tu<0.25fTcr Tu<0.25fTcr

j) Since Torsional affect on Girder is mainly caused by the Torsional Moments at Support Position & the Calculated

iv)of AASHTO-LRFD-5.8.3.6:

a) N-mm

b) 1030000

2000 200 2000 mm

350 mm,

c) 113.097

d) q

e) s mm

v)to Equ-5.8.3.6.2-2 & Equ-5.8.3.6.2-4:

a) Under Combined Shear & Torsion the Factored Shearing Force at a Section is 1,180.952 kN

1180.952*10^3 N

b) 4,156.978

0 . Where for both Equation; 4.156978

c) N

d) 4,300.000 mm

2000 50 12 mm

350 38 mm.

e) 494,656

Value of Torsional Monent, Tu

Value of 0.25fTcr

Statu between Tu & 0.25fTcr

Factored Torsional Moments at Support Position, Tu > 0.25fTcr, thus the Girder is Safe in respect of Tortional affect.

Computation of Nominal Torsional Resistance (Tn) Subject to Combined Shear & Torsion under Provisions

According to AASHTO-LRFD-5.8.3.6 the is Nominal Torsional Resistance of a Tn

Section, Tn = 2AoAtfycotq/s in N-mm. In which;

Ao is Area enclosed by Shear Flow Path. In T-Girder Ao = bhf + bWeb(hGir-hf), Ao mm2

Here, b = mm, hf = mm, hGir. =

& bWeb =

At is X-Sectional Area of One Ledge closed Transverse Torsional Reinforcement At mm2

in mm2. Here At = Av/2 (Av is X-Sectional Area of 2-Leg Transverse/Shear Reinforcement).

q is Angle (0) of Crack determined under provisions of AASHTO-LRFD-5.8.3.4. (O)

with necessary modifactions as required for v & Vu

s is Spacing of Transverse Torsional Reinforcement in mm.

Compution of Factored Shearing Force Vu & Shearing Stress vu for Combined Shear & Torsion accodring

Vu-S&T

Vu-S&T = Ö(Vu2+(0.9phTu/2Ao)2) ; Equ-5.8.3.6.2-2.

For Normal Section vu = Ö((Vu - fVp)/fbvdv)2 + (Tuph/fAoh)2) ; Equ-5.8.3.6.2-2 vu kN/m2

having Vp = N/mm2

Vu is Factored Shearing Force for the Section in N due to Applied Loads. Vu

ph is Parimeter of the Center Line of Closed Transverse Torsion Reinforcement in ph

mm. Here ph = 2*((hGir.-2CCov-B&T - f12) + (bWeb-2CCov-Side- f12)), in which

i) hGir = mm. ii) CCover-B&T = mm,iii) f12 =

iv) bv/bWeb= mm. v) CCov-side =

Aoh is the Area Enclosed by Center Line of Exterior Closed Transverse Torsional Aoh mm2

Reinforcement in mm2. Here Aoh = (hGir.-2CCover-B&T - f12)*(bWeb-2CCover-Side- f12)


g) N-mm

h) f 0.90

i)

Table-4. Showing the values of Mu, Vu,Tu, Vu-S&T, vu & other Items related to Torsion.

Location Length of Factored Factored Calculated Calculated Effective Factored Calculated Calculated

of Section Segment Moment Shear value of value of Moment Torsional value of value of

from from the for the for the Effective value Moment

Support Earlier Section. Section. Shear Depth of Section. of Section. of Section.

Section

mm kN-m kN mm kN kN-m kN-m kN

At Support 0.000 0.00 860.766 1711.800 1473.459 1473.459 430.383 1180.952 4.157

375.000 278.14 834.197 1711.800 1427.978 1427.978 417.098 1144.499 4.029

2675.000 3,003.20 644.670 1694.520 1092.406 3003.199 322.335 884.473 3.113

3050.000 5,131.01 407.258 1666.543 678.713 5131.011 203.629 558.749 1.967

3050.000 6,318.42 191.162 1614.600 308.650 6318.423 95.581 262.270 0.923

3050.000 6,630.45 -46.250 1627.400 -75.267 6630.449 -46.382 98.649 0.448

vi)provisions of Equation 5.8.3.4.2-1 :

a)

b) 2.901

c)

Torsion.

Location Length of Calculated Factored Provided Provided Provided Calculated Effective Calculated

of Section Segment value of Moment Steel Area Steel Area Total Steel value of Moment value of

from from the Effective for the for Flexural of S&T Bar Area for the value

Support Earlier Shear Depth Section. Main Bars on Top Face Section of Section. with Least

Section the Min.

mm mm kN-m kN kN-m Torsion Bar

At Support 0.000 1711.800 0.00 6433.982 1608.495 8042.477 1180.952 1473.459 1.000

375.000 1711.800 278.135 6433.982 1608.495 8042.477 1144.499 1427.978 0.969

2675.000 1694.520 3003.199 8042.477 1608.495 9650.973 884.473 3003.199 0.950

Tu is Factored Torsional Moment in N-mm for the Section. Tu

f is Resistance Factor for Torsion according to AASHTO-LRFD-5.5.4.2

Table Showing the values of Mu, Vu,Tu, Vu-S&T, vu & other Items related to Torsion at different Section of Girder.

Vudv Vu-S&T vu

MuEff

(Mu) (Vu) (dv) (Vudv) (MuEff) (Tu)N/mm2

L0.375m

L1/8

L1/4

L3/8

L1/2


For Nonprestressing RCC Girder values of Components Nu, Vp, Asp, fpo, Ep etc. are = 0. Substituting Vu by Vu-S-T

Equation for ex = (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/2(EsAs + EpAps) stands to

ex = (Mu/dv + 0.5Vu-S&Tcotq )/2EsAs

Considering the Initial value of ex = 0.001at Support Position & with the Effective cotq

Moment Mu-Eff. from the Equation value of Cotq = (ex2EsAs - Mu/dv)/0.5Vu

Having cotq = 1.718, Calculated values of exx1000 at Different Locations are shown in the under mentioned Table-6:

Table-6. Showing Calculated values of exx1000 at Different Locations Mu-Eff,, Vu-S&T, & other Items related to

Vu-S&T exx1000

MuEff

(dv) (Mu) (As-Main) (As-S&T) (As-Total) (MuEff)mm2 mm2 mm2

L0.375m

L1/8


3050.000 1666.543 5131.011 11259.468 1608.495 12867.964 558.749 5131.011 0.864

3050.000 1614.600 6318.423 16084.954 1608.495 17693.450 262.270 6318.423 0.667

3050.000 1627.400 6630.449 14476.459 1608.495 16084.954 98.649 6630.449 0.728

i)Locations is not essential.

j)

Equation No. 5.8.3.4.2-4 & Table -5.8.3.4.2-2.

viii) provisions of Equation 5.8.3.4.2-2 :

a)

b) 2.901

c) -

ix)

a)

b) - mmReinforcements on Vertical Faces (as Shrinkage & Tempeture Reinforcement)

201.062

- mm

s 232.857 mm

- 0.003dvs>As s-Applicable

L1/4

L3/8

L1/2

Since at L/2 Distance from Support the Shear Forces are of (-) ve value, thus calculated value of ex-x1000 these

Since in all Sections from Support to L/2 the values of es-x1000 ≤ 1.00, thus values of q & b for the Sections can be

obtain from Table-5.8.3.4.2-1 in respect of values of vc/f/c. For Sections having the es-x1000>1.00, for those Cases

values of q & b can obtain in respect of value of Crack Spacing Parameter sxe, using the Equation .No.5.8.3.4.2-2,

Values of esx1000 with Less than Min. Transverse/Shear Reinforcement under AASHTO-LRFD-5.8.2.5 & the

Since in RCC Girder the values for Prestressing Components Nu, Vp, Asp, fpo, Ep etc. are = 0, thus equation

ex= (Mu/dv+0.5Nu + 0.5(Vu -Vp)cotq - Apsfpo)/(EsAs + EpAps) stands to

ex = (Mu/dv + 0.5Vucotq )/EsAs

Considering the Initial value of ex = 0.002 at Support Position & the Effective cotq

Moment Mu-Eff. From the Equation Cotq = (exEsAs - Mu/dv)/0.5Vu

Values of exx1000 at Location = (Mu/dv + 0.5Vucotq )/EsAs ex-*1000

Computation of values of Crack Spacing Parameter, sxe for Section at L/8 :

The Crack Spacing Parameter of a Section, sxe = sx(35/(ag+16)) £ 2000mm, In which;

sx = the Lesser value of either dv or the Spacing of Longitudinal Crack Control sx

having Steel Area in a Horizontal Reinforcement Layer if As > 0.003bvs. Here;

i) As is Steel Area of 2nos Shrinkage & Tempeture Reinforcement Bars on As mm2

Opposite Vertical Faces of T-Girder in same Horizontal Layer,

= As-S&T-Hori. = 2*pDBar2/4 = 2*Af-16

ii) dv is Effective Shear Depth of Tensil Reinforcement for the Section dv

iii) s is Spacing of Longitudinal Bars as Shrinkage & Tempeture Reinforcement on Vertical Faces of T-Girder.

iv) Thus Computed value of 0.003bvs for a Section 0.003dvs mm2

v) For Computed value of 0.003bvs at Section > As, the value of sx = dv


c) 20 mm

d) - mm

f)

x)

a)

Location of Section From Referece Value of Value of Support of Table q

a) At Support Position 5.8.3.4.2.-1 0.198 1.000 NA 36.400 1.356

5.8.3.4.2.-1 0.192 0.969 NA 36.400 1.356

5.8.3.4.2.-1 0.148 0.950 NA 36.400 1.356

5.8.3.4.2.-1 0.094 0.864 NA 36.400 1.356

5.8.3.4.2.-1 0.044 0.667 NA 33.700 1.499

5.8.3.4.2.-2 0.021 0.728 NA 33.700 1.499

xi) of AASHTO-LRFD-5.8.3.6

a)

b)

c)

d)

Location Calculated Calculated Factored Calculated Calculated Provided Provided Status Status

of Section value of Factored Torsional Nominal value of Steel Area Spacing of between between

from Torsional Resistance Torsional of Torsional Torsional

Support Moment Resistance Bars Bars

(s)kN-m kN kN mm

At Support 1.356 430.383 666.3235 740.359 1030000.000 113.097 175.000 Tu<Tr Tu<Tn

1.356 417.098 666.323 740.359 1030000.000 113.097 175.000 Tu<Tr Tu<Tn

1.356 322.335 666.323 740.359 1030000.000 113.097 175.000 Tu<Tr Tu<Tn

1.356 203.629 583.033 647.815 1030000.000 113.097 200.000 Tu<Tr Tu<Tn

ag is Max. Aggragate size for Concrete = 20mm ag

Thus Computed value of sxe = sx(35/(ag+16)) for a Section sxe

For value of ex-*1000 > 1.00 & the value of sxe at a Sections should Computed from the Respective Tablel.

Computation of Values of q from AASHTO-LRFD'sTable -5.8.3.4.2-1. & Table -5.8.3.4.2-2. against the

Respective Calculated values of esx1000, Ratio vc/f/c & sxe.:

Table-7 for Values of q & Cotq at Different Location of Girder.

vu/f/c exx1000 sxe

Cotq

b) At L0.375mDistance of Support c) At L/8 Distance of Support d) At L/4 Distance of Support e) At 3L/8 Distance of Support f) At L/2 Distance of Support

Computation of Nominal Torsional Resistance (Tn) Subject to Combined Shear & Torsion under Provisions

According to AASHTO-LRFD-5.8.2.1 the Factored Torsional Resistance of a Section, Tr = fTn in N-mm.

According to AASHTO-LRFD-5.8.3.6 the Nominal Torsional Resistance of a Section, Tn = 2AoAtfycotq/s in N-mm.

The Computed values for Factored Torsional Moment Tu, Factored Torsional Resistance Tr & the Nominal Torsional

Resistance Tn in respect of provided Steel Area for Torsional Reinforcement Combined with Shear Reinforcement At,

Spacing of Combined with Shear & Torsional Reinforcement s, Computed values of Ao the Area enclosed by Shear Flow Path & Cotq at different Section along with respective Status are shown in the under mentioned Table-8.

Table-8. Showing the values of Tu, Tr,Tn, At, Ao, Cotq , s & Status of different events related to Torsion.

Cotq Ao Tu &Tr Tu &Tn

(Tu) (Tr) (Tn) (At)mm2 mm2

L0.375m

L1/8

L1/4


1.499 95.581 644.531 716.146 1030000.000 113.097 200.000 Tu<Tr Tu<Tn

1.499 -46.382 644.531 716.146 1030000.000 113.097 200.000 Tu<Tr Tu<Tn

e)

xii) Checking in respect of Longitudinal Reinforcements (Tensile Reinforcements) Provided for Girder under Provision of AASHTO-LRFD-5.8.3.6.3 :


Where;

b) Variable

c) 410.000 MPa

d) -

e) - MPa

f) Variable N-mm


h) Variable mm

i) Variable N

j) Variable N

k) - N


m) 0.90 AASHTO-LRFD-5.5.4.2.

L3/8

L1/2

At all Section the Factored Torsional Moment Tu <Tr & the Factored Torsional Resistance& also Tu < Tn, Nominal Torsional Resistance, thus the design is in respect of Combind Shear & Torsion.

proportationed to Satisfy Equation-5.8.3.6.3-1. (This is related to Equation-5.8.3.5-1.with some modifications).

Asfy + Apsfps ³ Mu/dvff + 0.5Nu/fc + cotqÖ((Vu/fv - 0.5Vs -Vp)2+(0.45phTu/2Aof)2) . (Equ-5.8.3.6.3-1).



















n) 0.90 AASHTO-LRFD-5.5.4.2,

o) 0.80 AASHTO-LRFD-5.5.4.2.

p) f 0.90 AASHTO-LRFD-5.5.4.2.

q)

r) Table- Showing Evalution of Equation-5.8.3.6.3-1 at Different Section of Girder & Status of Results.

Section Factored Calculted Calculted Status

from Effective Provided Factored Factored Shearing Torsional Value of R/H Part of Equation

Bearing Shear Steel Area Moment Shearing Resistance Moment L/H Part of the Satisfy/

Center of Depth of Section Force of Stirrups Equation Does

Girder. mm kN-m kN kN kN-m kN kN Not Satisfy

a)At Support 1711.800 8042.477 0.000 860.766 1230.434 430.383 3297.416 765.090 Satisfyed

1711.800 8042.477 278.135 834.197 1230.434 417.098 3297.416 906.719 Satisfyed

1694.520 9650.973 3003.199 644.670 1230.434 322.335 3956.899 2445.681 Satisfyed

1666.543 12867.964 5131.011 407.258 1065.762 203.629 5275.865 3783.971 Satisfyed

1614.600 17693.450 6318.423 191.162 1158.726 95.581 7254.314 4863.915 Satisfyed

1627.400 16084.954 6630.449 326.980 1122.611 163.490 6594.831 4881.476 Satisfyed

j) Since at all Section the requirments of Equation are being Satisfied, thus provision of Transverse/Shear & Torsional Reinforcements for Girder is OK.

7 Checking against Deflection & Camber :

i) Flexural Deformation of Components Due to Deflection & Camber, (AASHTO-LRFD-5.7.3.6.2) :

a)

b)

c)

Component.

ii) Instantaneous Deflections at Mid Span against Individual Applied Forces :



f is Resistance Factor for Torsional Force of Reinforced Concrete according to


Equ-5.8.3.6.3-1. Stands to, Asfy ³ Mu/dvff + cotqÖ((Vu/fv - 0.5Vs)2 + (0.45phTu/2Aof)2).

dv. As. Mu Vu Vs

(Tu) Asfy

mm2

b) At L0.375m.

c) At L/8.

d) At L/4.

e) At 3L/4.

f) At L/2.

Note ;- The (-) ve Shear & respective Torsional Moment values at Section L/2, are being cosidered as (+) ve value.

Deformation/ Deflection of Structural Components in respect of Dead & Live Load, Creep, Shrinkage, Thermal Changes, Settlement & Prestressing should be considered according to provision of AASHTO-LRFD-2.5.2.6.

In Calculation of Deflection & Chamber Dead Load, Live Load, Prestressing, Erection Load, Concrete Creep & Shrinkage and Steel Relaxation should be Considered. For Determining Deflection & Camber the Provisionsof AASHTO-LRFD-4.5.2.1; 4.5.2.2 & 5.9.5.5. are applicable.

Computation of Instantaneous Deflection should be based on Modulus of Elasticity of Conncrete-Ec as mentioned

in AASHTO-LRFD-5.4.2.4. & either the Gross Moment of Inertia-Ig or the Effective Moment of Inertia-Ie of the


a) 20.612 mm

b) 4.451 mm

c) 0.000 mm

d) 2.288 mm

e) 0.502 mm

f) 0.878 mm

g) 5.725 mm

h) 6.424 mm

i) Total Instantaneous Deflection at Mid Span against Individual Applied Forces 40.881 mm

iii) Instantaneous Deflections at Mid Span against Calculated Moments for Applied Forces under Service Limit State (WSD) :

a) 23.633 mm

b) 39.823 mm

iv) Instantaneous Deflections at Mid Span against Calculated Moments for Applied Forces under Strength Limit State (USD) :

a) 30.191 mm

b) 66.212 mm

v) Calculation of Long Term Deflections :

a) 0.269

2.695E+11

b) 3.000

Deflection due to Uniformely Distributed Factored Dead Loads (DL) DInst-UDL-DL

Deflection due to Uniformely Distributed Factored Live Lane Loads (LL) DInst-UDL-LL

Deflection due to Factored Concentrated Dead Loads (DL) for 1st & 5th DInst-CDL-X-1&5.

Cross Girder.

Deflection due to Factored Concentrated Dead Loads (DL) for 2nd & 4th DInst-CDL-X-2&4.

Cross Girder.

Deflection due to Factored Concentrated Dead Loads (DL) for 3rd DInst-CDL-X-3.

Cross Girder = KCL-DL-3-X-Gir-MidMCDL-X-3.L2/EcIe.

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Front.

Front Wheel

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Mid.

Middle Wheel

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Rear.

Rear Wheel = KCL-LL-Wheel-Rear(X<L/2)MCLL-W-Rear.L2/EcIe.

DInst.-Total

Instantaneous Deflection Interior Beam at Mid Span due to Dead Load (DL) DInst-DL-WSD

Instantaneous Deflection Interior Beam at Mid Span due to Total of DL + LL DInst-Total-WSD

Instantaneous Deflection Interior Beam at Mid Span due to Dead Load (DL) DInst-DL-USD

Instantaneous Deflection Interior Beam at Mid Span due to Total of DL + LL DInst-Total-USD

The Instantaneous Deflections are being Calculated Based on Ie, the Effective Ie m4

Moment of Inertia Computed according to Equation 5.7.3.6.2-1; Article 5.7.3.6.2. mm4

The Instantaneous Deflection is Based on Ie, thus Multiplying Factor l = 3.000 l-Ie.


c) 119.469 mm

vi) Computation of Rotations at Support Position of Girder due to Different Applied Loads :

a) 2.449E-15 Radian.

b) 1.588E-15 Radian.

c) Rotation on Support Position for Unfactored Live Load only under Service State 1.59E-15 Radian. of Design (WSD) due

vii) Limit of Deflection & Camber under AASHTO-LRFD :

a)

b) Deflections can consider According to following Limit for Steel, Aluminum and/or Concrete Constructions with different Combination of Loads are the Followings :

L/800 30.500 mm

L/1000 24.400 mm

L/300 81.333 mm

L/1000 65.067 mm

c) Since the calculated Deflections & Rotations are more or less within the Limit, thus Structure is Safe.

8 Flexural Design of Diaphragm/Cross-Girder :

i) Design Phenomena :

a)

b) The Cross-Girders at Interior Positions will have to face the Max. Moments & Shearing Forces caused by the applied

one based on applied Moments & Shears. Since the Bridge Deck Slab is integral Part of Girders, thus the Design of

c)

= 0.41 m

2.65 m

Calculated value of Long Term Deflectiom at Mid Span = lIe*DInst-Acceptable. DLong.

Rotation on Support Position under Strength Limit State of Design (USD) due qGir-USD

to Total Applied Loads (DL + LL)

Rotation on Support Position under Service Limit State of Design (WSD) due q-Gir-WSD

to Total Applied Loads (DL + LL)

q-Gir-LL

According to AASHTO-LRFD-2.5.2.6.2 Deflection & Camber is Optional Item for RCC Bridge Structure.

i) For general Vehicle Load, D = L (Span Length)/800

ii) For Vehicle and/or Padestrain Loads, D = L (Span Length)/1000

iii) For Vehicle Load on Cantilever, D= L (Span Length)/300

iv) For Vehicle and/or Padestrain Loads on Cantilever, D = L (Span Length)/375

The Flexural of Girders will be according to AASHTO LRFD or Ultimate Strength Design (USD) Procedures.

Loads (DL& LL), thus it is require to conduct the Flexural Design of an Interior Cross-Girder as Typical one based on

Girders will be under T-Beam if the Provisions in these Respect Satisfy, otherwise Designee will be under Provisionsfor the Rectangular Beam.

Calculation of Effective Flange Width of T-Girder under AASHTO LRFD-4.6.2.6 (4.6.2.6.1) as least Dimention of :

i) One-quarter of Effective Span Length = 1/4*SL-X-Gir

ii) 12.0 times average Depth of Slab + Greater Thickness of Web = 12*tSlab + bX-Gir. =iii) One-half the Width of Girder Top Flange (It is not req. as there is no Addl. Top Flange)


= 6.25 m

d) Since One-quarter of Effective Span Length is the Least one, thus the 0.41 m

ii) Dimensional Data of Cross-Girder :

a) Span Length of Cross-Girder (Clear distance between Main Girder Faces) 1.650 m

b) Thickness of Deck Slab 0.200 m

c) Thickness of Wearing Course 0.075 m

d) Number of Cross Girders 5.000 nos

e) Depth of Cross Girders (Including Slab as T-Girder) 1.900 m

f) Width of Cross-Girder Web 0.250 m

g) Width of Main-Girder Web 0.350 m

h) C/C Distance Between Main Girders 2.000 m

i) Flenge Width of Cross-Girder 0.413 m

j) C/C Distance in between Cross-Girders in Longitudinal Direction 6.250 m

k) Filets : i) X-Girder in Vertical Direction 0.075 m

ii) X-Girder in Horizontal Direction 0.075 m

iii) Sketch Diagram of Cross-Girder T-Beam :

0.413

0.200 m

1.900 m0.010 m

0.250 m

iv) Computation of Daed Loads on Cross-Girders :

a) 10.335 kN/m

b) 1.980 kN/m

c) 0.557 kN/m

d) 12.872 kN/m

iv) The average Spacing of Adjacent Beams/Girders = CD-X-Gir.

bFl-Gir.

Flange Width of Cross-Girders, bFl-X-Gir = 1.450m

SL-X-Gir.

tSlab.

tWC

NX-Gir.

hX-Gir.

bX-Web.

bMain-Web.

C/CD-Gir.

bFln-X-Gir.

CD-X-Gir.

FX-Gir-V.

FX-Gir-H.

bFln-X-Gir

hf-X-Gir =

hX-Gir. =d =

bX-Web =

Dead Load due Self Weight (Excluding Slab & W/C but including Fillets) in kN DLX-Gir-Self

for per Meter Span Length = wc*((hX-Gir - tSlab)*bX-Web + 2*0.5*FX-Gir-V*FX-Gir-H)

Dead Load from Slab (Within Flange Width) in kN for per Meter Span Length DLX-Gir-Slab.

= gc*bFln-X-Gir.*tSlab

Dead Load from Wearing Course (Within Flange Width) in kN for per Meter DLX-Gir-WC.

Span Length = gWC*bFln-X-Gir.*tWC

Summation of Dead Loads of Cross-Girder for per meter Length åDLX-Gir.


v) Computation of Live Loads on Cross-Girders :

a) Live Loads on Cross-Girders will be from Moving Truck due to its Wheel Loads & Lane Loads Over the Deck Slab.Truck Load will produce Max. Reaction if either of its Rear or Middle Wheel take position directly over Cross-Girder. Since the Main Girders have 2.000m C/C Spacing, thus only One no. of Wheel can took position over Cross-Girder.Whereas Lane Load will produce Uniformly Distributed Reaction for Cross-Girder having its Intensity of action within the dimensions that of Cross-Girder Span & Flange Width.

b) Concentrated Live Load Reaction due to Rear/Middle Single Wheel from 72.500 kN

c) 1.279 kN/m

vi) Factored Dead Load for per meter Length of Cross-Girder:

a) 15.394 kN/m

b) 0.835 kN/m

c) 16.229 kN/m

vii) Factored Live Loads of Cross-Girder :

a)

168.744 kN

b) 2.238 kN/m

viii) Computation of Shearing Forces at Different Location due to Factored Dead Load & Live Load :

a) The Cross-Girders are Structurally Integral Components of Bridge Main Girders having both side Fixed Ends and will have Shearing Forces accordingly. The Dead Loads & Lane Live Loads are Uniformly Distributed Loads over the Cross-Girder Span. Whereas Wheel Live Load is a Concentrated one having scope & attitude of changing position over the Cross-Girder Span.

viii-i) Shearing Forces due to Unifomly Distributed Factored Dead Loads (FDL) :

a) Shearing Forces on Cross-Girder at Faces of Main Girder due to Unifomly 13.389 kN

b) Shearing Forces at Middle of Cross-Girder Span due to Unifomly - kN

LLX-Gir.-Wheel.

Truck in kN = LLRSW-Load or LLMSW-Load

Uniformly Distributed Live Load due to Loan Load Reaction in kN per Meter LLX-Gir-Lane.

Span Length = LLLane/3*bFln-X-Gir.

Factored Dead Load of Cross-Girder due to Self & Slab in kN/m FDLX-Gir+Slab.

= gDC*(DLX-Gir-Self + DLX-GirSlab)

Factored Dead Load of Cross-Girder due to Wearing Course in kN/m FDLX-Gir-WC

= gDW*DLX-Gir-WC

Summation of Factored DL of Cross-Girder for per meter Length åFDLX-Gir.

Factored LL of Cross-Girder due to Wheel Load in kN;

= mgLL-Truck*IM*LLX-Gir-Wheel FLLX-Gir.-Wheel.

Factored LL of Cross-Girder due to Lane Load in kN/m FLLX-Gir-Lane.

= mgLL-Lane*LLX-Gir-Lane.

VX-Gir-DL-Face

Distributed Factored Dead Load of Girder = åFDLX-Gir. * SL-X-Gir./2.

VX-Gir-DL-Center


viii-ii) Shearing Forces due to Unifomly Distributed Factored Live Lane Loads (FLLL) :

a) Shearing Forces on Cross-Girder at Faces of Main Girder due to Unifomly 1.846 kN

b) Shearing Forces at Middle of Cross-Girder Span due to Unifomly - kN

viii-iii) Shearing Forces due to Factored Concentrated Wheel Loads Posioned at Mid of Span(FLWL-Mid) :

a) Shearing Forces on Cross-Girder at Faces of Main Girder due to 84.372 kNFactored Live Wheel Load (Concentrated) having its position at Center

b) Shearing Forces on Cross-Girder at Center of Cross Girder due to 84.372 kNFactored Live Wheel Load (Concentrated) having its position at Center

viii-iv) Shearing Forces due to Factored Concentrated Wheel Loads Posioned at Face of Span (FLWL-Mid) :

a) Shearing Forces on Cross-Girder at Faces of Main Girder due to 168.744 kNFactored Live Wheel Load (Concentrated) having its position at one Face

b) Shearing Forces on Cross-Girder at Center of Cross Girder due to 84.372 kNFactored Live Wheel Load (Concentrated) having its position at one Face

h) Table-1 Showing the Shearing Forces at different Locations of Cross-Girder due to Dead & Live Loads :

Sl. No. Type of Loading Loactions/ Position At Main At Cross

of Load Application Girder Face Girder Mid.

Unit kN kNi). Uniformly Distributed Throught the Span 13.389 0.000

Dead Load (DL) Length ii). Uniformly Distributed Throught the Span 1.846 0.000

Live Lane Load (LL) Length iii). Concentrated Wheel At Center of the 84.372 84.372

Live Load (LL) Cross-Girder Span iv). Concentrated Wheel At Main Girder Face 168.744 84.372

Live Load (LL)Total Shearing Forces due to applied Loads at 268.351 168.744 Different Loactions of Cross-Girder.

Distributed Factored Dead Load of Girder = VX-Gir-DL-Face - åFDLX-Gir. * SL-X-Gir./2 .

VX-Gir-Lane-Face

Distributed Factored Live Lane Load on Girder = FLLX-Gir-Lane.* SL-X-Gir./2.

VX-Gir-Lane-Center

Distributed Factored Live Lane Load on Girder = VX-Gir-X-Gir-Lane. - åFLLX-Gir.-Lane. * SL-X-Gir./2 .

VX-Gir-Face-Wheel-Mid

of Span will be = FLLX-Gir-Wheel. /2

VX-Gir-Center-Wheel-Mid

of Span will be = FLLX-Gir-Wheel. /2 (The (+) ve. value only).

VX-Gir-Face-Wheel-Face

of Span will be = FLLX-Gir-Wheel.

VX-Gir-Center-Wheel-Face

of Span will be = 1/2*FLLX-Gir-Wheel.


ix) Computation of Moments at Different Location due to Unfactored Dead Load, Factored Dead Load & Live Loads :

a) The Cross-Girders are Structurally Integral Components of Bridge Main Girders having both side Fixed Ends and will have Moments due to Loads accordingly. The Dead Loads & Lane Live Loads are Uniformly Distributed Loads over the Cross-Girder Span. Whereas Wheel Live Load is a Concentrated one having scope & attitude of changing position over the Cross-Girder Span.

b) (+) ve. Moment at Center of Cross-Girder Span due to Uniformly Distributed 2.503 kN-m

c) (-) ve. Moment at Face of Main-Girder Span due to Uniformly Distributed 2.920 kN-m

d) (+) ve. Moment at Center of Cross-Girder Span due to Uniformly Distributed 3.156 kN-m

e) (-) ve. Moment at Face of Main-Girder Span due to Uniformly Distributed 3.682 kN-m

f) (+) ve. Moment at Center of Cross-Girder Span due to Uniformly Distributed 0.435 kN-m

g) (-) v. Moment at Face of Main-Girder Span due to Uniformly Distributed 0.508 kN-m

h) (+) ve. Moment at Cross-Girder Center due to Concentrated Factored 34.803 kN-m

i) (-) ve. Moment at Main-Girder Face due to Concentrated Factored 34.803 kN

j) (+) ve. Moment at Cross-Girder Center due to Concentrated Factored 139.214 kN-m

k) (-) ve. Moment at Main-Girder Face due to the Concentrated Factored Live 33.163 kN-m

0.175 m

l) Table-2 Showing the Moments at different Locations of Cross-Girder due to Factored Dead & Live Loads :

(+) MUF-DL

Unfactored Dead Load (DL), M = åDLX-Gir. * SL-X-Gir.2/14.

(-) MUF-DL

Unfactored Dead Load (DL); M = åDLX-Gir. * SL-X-Gir.2/12.

(+) MF-DL

Factored Dead Load (FDL), M = åFDLX-Gir. * SL-X-Gir.2/14.

(-) MF-DL

Factored Dead Load (FDL); M = åFDLX-Gir. * SL-X-Gir.2/12.

(+) MF-LLL

Factored Live Lane Load (FLLL); M = åDLX-Gir.-lane * SL-X-Gir.2/14.

(-) MF-LLL

Factored Live Lane Load (FLLL); M = åDLX-Gir.-Lane * SL-X-Gir.2/12.

(+)MF-Wheel-Mid

Live Wheel Load (FWLL) having its position at Center of Span will be

M = FLLX-Gir-Wheel. *SL-X-Gir./8

(-)MF-Wheel-Mid

Live Wheel Load (FWLL) having its position at Center of Span will be

M = FLLX-Gir-Wheel. *SL-X-Gir./8

(+)MF-Wheel-Face

Live Wheel Load (FWLL) having its position at Face of Main Girder will be

M = FLLX-Gir-Wheel.*SL-X-Gir./2

(-)MF-Wheel-Face

Wheel Load (FWLL) having its position at a Distance a = bMain-Web/2 from the

Face of Main Girder will be M = a*(a-1)2*FLLX-Gir-Wheel.*SL-X-Gir.

Here bMain-Web is Width of Main-Girder Web; thus a =


Sl. No. Type of Load Application Position (+) Moments at (-) Moments at . Location of Load Cross-Girder Center Main Girder Face

i). Uniformly Distributed Throught the Span 0.435 kN-m 3.682 kN-m Length

ii). Uniformly Distributed Throught the Span 0.435 kN-m 0.508 kN-m Length

iii). Concentrated Wheel At Center of the 34.803 kN-m 34.803 kN-m Cross-Girder Span

iv). Concentrated Wheel At Main Girder Face 139.214 kN-m 33.163 kN-m

Total Moments due to applied Loadsat different 174.887 kN-m 72.156 kN-m Loactions of Cross-Girder.

x) Factored Flexural Resistance for Prestressed or RCC Structural Components (AASHTO-LRFD-5.7.3.2.1):

a) N-mm

N-mm

f

b)

c) For a Nonprestressing Structural Component of Rectangular Elements having Singly Reinforced, , at any Section the

xi) Limits For Manimum Reinforcement, (AASHTO-LRFD-5.7.3.3.2) :

a) For Section of a Flexural Component having Prestressed & Nonprestressed Tensile Reinforcements or only with

b) N-mmwhere;

- at Extreme Fiber where Tensile Stress is caused by Externally Applied Forces

N-mm

-

Dead Load (DL)

Live Lane Load (LL)

Live Load (LL)

Live Load (LL)




For a Nonprestressing Structural Component either of I or T Section having Flenge & Web Elements, at any Section

the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

Nominal Resistance, Mn = Asfy(de-a/2)

Nonprestressed Tensile Reinforcements should have Minimum Resisting Moment Mr ³ 1.2*Mcr or 1.33 Times the Calculated Factored Moment for the Section Based on AASHTO-LRFD-3.4.1-Table-3.4.1-1, which one is less.










150,416,666.667

0.150417

150.417/10^3

2.887

c) 434.256 kN-m

434256329.29 N-mm

d) N-mm

e) N-mm

f) N-mm

g)

Location of Value of Value of Actuat Acceptable Factored Allowable Min. Maximum

Section Unfactored Cracking Cracking of Cracking Moment as of Factored Resisting Flexurakfrom Dead Load As per Moment Moment Moment per Art.-3.4.1 Moment Moment Moment

Support Moment Equation Value (Tab-3.4.1-1) M-1

5.7.3.3.2-1 M (1.33*M)kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

At Middle 2.503 434.256 434.256 434.256 521.108 174.887 232.600 232.600 232.600

of SpanAt Fcae 2.920 434.256 434.256 434.256 521.108 72.156 95.968 95.968 95.968

of Span

xii)

a) 174.887 kN-m/m

174.887*10^6 N-mm/m

Moment value Reinforcement will be on Bottom Surface of Cross Girder. 232.600 kN-m/m 232.600*10^6 N-mm

b) 232.600 kN-M

232.600*10^6 N-mm

c) 50 mm

50 mm

d) 20 mm

e) 314.159



For the Rectangular RCC Girder Section value of Snc = (bX-WebhX-Gir3/12)/(hX-Git/2) m3








1.2 Times 1.33 Times

Mcr-1

Mcr Mcr Mr Mu

MDL-UF Sncfr (Mcr-1£Sncfr) (1.2*Mcr) (1.2Mcr£1.33M) (M ³ Mr)

Flexural Design of Cross-Girder with (+) ve. Moment value at Middle of Span for Reiforcement on Bottom :

The Calculated (+)ve Moment values at Middle Positions of Cross Girder (+)MX-Gir-Mid-USD

MX-Gir-Mid-USD < Mr, the Required Minimum Flexural Strength Moment.For (+) ve

Mr

Since MX-Gir-Mid-USD< Mr, the Allowable Minimum Moment for the Section, Mu

thus Mr is the Design Moment MU.

Let the Clear Cover at Bottom Surface of Cross-Girder, C-Cov.Bot. = 50mm, C-Cov-Bot.

Let the Clear Cover at Top of Cross-Girder, C-Cov.Top = 50mm, C-Cov-Top.

Let the Main Reinforcements are 20f Bars in 1 Layer, DBar



f) 32 mm

g) 10 mmof Vertical Stirrups.

h) Thus Effective Depth of Reinforcements from Top of X-Girder up to Center of the 1,830.000 mm

h) 0.022

i) 0.016

xiii) Checking's Whether the Cross Girder would Designed as T-Beam or Rectangular Beam Provisions :

a) According to Ultimate Stressed Design Provisions a Rectangular having Flange with Reasonable Thickness on its

b)Rectangular Beam.

c) a 17.344 mm

a<hFln

d)

xix) Provision of Tensile Reinforcements on Bottom Surface against Calculated Positive Moments :

a) 474,737.910


d) 1,511.138 nos.

e) 4 nos

f) 1,256.637

h) 69.973 mm

a<hFln Satisfied




de-pro-Top

Provided Reinforcements de = (hX-Gir - C-Cov-Top -DStri- 0.5DBar)

Balanced Steel Ratio for Grider Section according to AASHTO-1996-8.16.2.2 rb.

rb.= 0.85*0.85*(f/c/fy)*{599.843/(599.843+fy))

Max. Steel Ratio, rMax = 0.75*rb. (AASHTO-1996-8.16.2.1) rMax

Top should be Designed as T-Beam if Depth of Equivalent Compression Block 'a' is Greater than Flange Thickness


For Rectangular Section having Provided Effective Depth de-pro-Bot; Beam Width

b and the Calculated Max. Factored Ultimate Moment (Positive Moment) MU;

the value Equivalent Compression Block a = de(1 - (1 - 2MU/0.85f/cbde

2)1/2)

Since the Calculated value of Equivalent Compression Block a < hFln; the thickness of X-Girder Flange, thus the Flexural Design of T-Girder will according to Provisions of Rectangular Beam.

With MU, Design Moment; b, Width of Rectangular Beam; de-pro-Bot, Provided As-req.Bot mm2


the Section; As = MU/[ffy(dasu-Bot - a/2)]

Number of 20f bers required = As-req-Bot/Af-20 NBar-req

Let Provide 4nos 20f bars in One Layer as Bottom Reinforcment of X-Girder. NBar-pro.

Provided Steel Area for the Section with 4nos.20f bars = Nbar-pro*Af-20 As-pro mm2


Rectangular Section of X-Girder = As-pro*fy/(0.85*f/c*b)

MResis


Mr>Mu Satisfied


k)

xx) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

a) 0.450

b) c 59.477 mm

c) 0.85

d) 0.033

e) c/de-pro<c/de-max. OK

xix)

a) 72.156 kN-m/m

72.156*10^6 N-mm/m

Moment value Reinforcement will be on Top Surface of Cross Girder. 95.968 kN-m/m 95.968*10^6 N-mm

b) 95.968 kN-M

95.968*10^6 N-mm

c) 50 mm

50 mm

d) 20 mm

e) 314.159

f) 32 mm

g) 10 mmof Vertical Stirrups.

h) Thus Effective Depth of Reinforcements from Top of X-Girder up to Center of the 1,830.000 mm



Since against Provided Steel Area; i) the Equivalent Compression Block 'a'< hFln, Thickness of Cross Girder Flange;

ii) the Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. TheAllowable Max. Steel Ratio; thus Provisions & Flexural Design for Tensile Reinforcement for Bottom Surface is OK.







Flexural Design of Cross-Girder with (-) ve. Moment value at Faces of Span for Reiforcement on Top :

The Calculated (-)ve Moment values at Support Positions of Cross Girder (-)MX-Gir-Sup-USD

MX-Gir-Sup-USD < Mr, the Required Minimum Flexural Strength Moment.For (-) ve

Mr

Since MX-Gir-Sup-USD< Mr, the Allowable Minimum Moment for the Section, Mu


Let the Clear Cover at Bottom Surface of Cross-Girder, C-Cov.Bot. = 50mm, C-Cov-Bot.

Let the Clear Cover at Top of Cross-Girder, C-Cov.Top = 50mm, C-Cov-Top.

Let the Main Reinforcements are 20f Bars in 1 Layer, DBar




de-pro-Top


j)

xx) Provision of Tensile Reinforcements on Top Surface against Calculated Negative Moments :

a) 474,737.910


d) 1,511.138 nos.

e) 4 nos

f) 1,256.637

h) 69.973 mm

a<hFln Satisfied


Mr>Mu Satisfied

j) 0.002 p-pro<J2486 Satisfied

k)

xxi) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

a) 0.450

b) c 59.477 mm

c) 0.85

d) 0.033


xxii) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2.1:

Provided Reinforcements de = (hX-Gir - C-Cov-Top -DStri- 0.5DBar)

Since it being already found that the Calculated value of Equivalent Compression Block a < hFln; the thickness of X-Girder Flange, thus the Flexural Design of T-Girder will according to Provisions of Rectangular Beam.

With MU, Design Moment; b, Width of Rectangular Beam; de-pro-Top, Provided As-req.Bot mm2


the Section; As = MU/[ffy(de-pro-Top - a/2)]

Number of 20f bers required = As-req-Bot/Af-20 NBar-req

Let Provide 4nos 20f bars in One Layer as Top Reinforcment of X-Girder. NBar-pro.

Provided Steel Area for the Section with 4nos.20f bars = Nbar-pro*Af-20 As-pro mm2


Rectangular Section of X-Girder = As-pro*fy/(0.85*f/c*b)

MResis



Since against Provided Steel Area; i) the Equivalent Compression Block 'a'< hFln, Thickness of Cross Girder Flange;

ii) the Developed Resisting Moment MResis > MU, the Design Moment & iii) the Provided Steel Ratio ppro < pMax. TheAllowable Max. Steel Ratio; thus Provision & Flexural Design for Tensile Reinforcement for Top Surface is OK.








a) 803.493 kN-m400.069*10^6 N-mm

b) 892.770 kN-m892.770*10^6 N-mm

c) f 0.90

d)

e)

f) 892.770 kN-mSteel Area against Factored (+) ve.Max. Moments at its Mid Span will have value of 892.770*10^6 N-mm

g) 174.887 kN-mCross-Girder. 174.887*10^6 N-mm

h) Mr>Mu

Satisfied

xxiii)

(Absulate Max. Moment at c.g. Point);

9 Shrinkage & Temperature Reinforcement in Longitudinal Direction on Vertical Faces of Cross-Girders :

a) Girders are Simply Supported Structure having the Longitudinal Flexural Reinforcements due to Moments on Bottom Surface & also on Top Surface (In case of Doubly Reinforced or under Shrinkage & Temperature Provision). There willbe also Lateral & Vertical Reinforcements on Bottom, Top & Vertiocal Surfaces under the provisiona of Shear & WebReinforcement. It also requires Longitudinal Reinforcement on its Vertical Faces, those can provide under Shrinkage

b) Let consider 1 (One) meter Strip Length of Girder for Calculation of Shrinkage 1.00 m& Temperature Reinforcement in Longitudinal Direction on Vertical Faces of Girder. 1,000 mm

c) 1.90 m 1,900 mm


Mn is Nominal Resistance Moment for the Section in N-mm Mn

f is Resistance Factor for Flexural in Tension of Reinforcement/Prestressing.



y(d/s-a/2) + 0.85f/



Section the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

For Fixed Supported & Single Reinforced T-Cross-Girder the with Provided Mn-Mid-Span


Calculated Factored (+) ve. Moment MU at Mid Span Maximum for T-Section of (+) MU



Since, i) The Value of Resisting Moment > Design Moment;

ii) The Calculated value of c/de £ 0.42 & iii) the Provided Stee Ratio for Rectangular Section of

T-Girder, rRec-pro < rMax. Balance Steel Ratio,

iii) The Computed Factored Flexural Resistance Mr > Mu the Actual Factored Moment at Mid Span

Thus Flexural Design of Reinforcements for the T-Girder Span is OK.


LGirder.

Let consider the Depth of Gross-Girder as Height = hX-Gir m hX-Gir.


d) 509.76 & Temperature Reinforcement for Structural Components with Thickness Less

e) 1900000.000

f) 16 mmFaces of Girder.

g) 201.062

h) Spacing of Shrinkage & Temperature Reinforcements is less of 3-times the 0.75 mm

0.75 mm or = 450 mm

i) 2.535 nos.Shrinkage & Temperature Reinforcements on Vertical Surface of Girder

j) 749.413 mmReinforcementon Vertical Faces of Girder for Calculated Steel Area

k) 4 nosReinforcement.

l) 316.667 mm

m) 804.248

n)

10 Design of Shear Reinforcement for Cross-Girder against Shearing Forces & Checking:

i)

a)

Table-2. Sum. of Max. Shear Forces Against All Applied Loads (DL & LL) on Interior Girder.

Locations of Searing Force On Main At 2d At Middle

Girder Distance of Cross

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T.-Hori. mm2

than 1200mm; As ³ 0.11Ag/fy .

Here Ag is Gross Area of Girder's Each Vertical Face = LGirder*hX-Gir. Ag-Vert. mm2



s-req-S&T.-Hori.


Number of 16f bars required against Calculated Steel Area as Horizontal NBar-req-S&T-Hori.

= As-req-Long./Af-16

Spacing of 16f bars as Horizontal Shrinkage & Temperature sCal-req-S&T-Hori.

= Af-16. *hX-Gir/As-req-Long.

Let Provide 5 nos. 16f as Longitudinal bars as Shrinkage & Temperature NBar-pro-S&T-Hori.

Spacing for provided 4 nos.16f bars as Longitudinal Shrinkage & spro-S&T-Hori.

Temperature on Vertical Faces of Girder = hX-Gir/(NBar-ro-S&T-Hori. +2)

Steel Area against Provided 4 nos. 16f bars as Longitudinal Shrinkage & As-pro-S&T-Hori. mm2

Temperature Reinforcement on Vertical Surfaces of Girder = Af-16*NBar-pro-S&T-Hori.

Since As-pro-S&T-Hori. > As-req-Long & spro-S&T-Hori.> s-req-S&T-Hori. thus the provisions of Longitudinal Reinforcement as Shrinkage & Temperature on Vertical Faces of Cross-Girders on both Faces are OK.


Table for Max. Shear Forces at Different Locations of Interior Girder due to Factored DL, Lane-LL & Wheel-LL :


Faces from Face GirderUnit of Shearing Forces kN kN kNTotal Shears on Each Point 268.351 168.744

b) Shearing Forces at Sopport Position of Girder 268.351 kN

c) Shearing Forces at Middle of Girder. 168.744 kN


a)

b) - Mpa

c) 250 mm

d) Variable mm


0.72h 1.368 mm

Variable mm

h 1.900 mm

e) f 0.90

f)

Location Length of Width of Depth of Effective Calculated Calculated Calculated Calculated

of Section Segment T-Girder T-Girder Depth of value of value of value of value of

from from the Web Section for 0.9de 0.72h Effective Shearing

Support Earlier Tensial Shear Depth Stress

Section (h)mm mm mm mm mm mm mm

At Faces 1.65 250.00 1.90 1830.00 1647.00 1.368 1647.00 0.724On Middle 0.83 250.00 1.90 1830.00 1647.00 1.368 1647.00 0.455

g) Shearing Stress due to Applied Factored Shearing Forces at Different Sections of Girder :

i) Shearing Stress due to Applied Shearing Force at Support Position of Girder 0.724

Vu-Face.

Vu-Mid.




bv is Width of T-Girder Web = bX-Web. mm, bv


0.9de



ii) h is Depth of T-Cross-Girder = hX-Gir mm,


Table for Computation of values of vu, de, bv, 0.9de , 0.72h & dv at different Section of Girder.

Table-3 Values of vu, de, bv, 0.9de , 0.72h & dv at different Section of Girder.

(bv) (de) (dv) (vu)N/mm2

vC-Face. N/mm2

= Vu-Face/fbvdv N/mm2


ii) Shearing Stress due to Applied Shearing Force at Middle of Cross-Girder Span 0.455


a)

b) N

c) N

d) 0.90


a)

b)

c)

d)

e)

f)

g)

a 90

h)

i) - N

v)

a)

vMid. N/mm2

= Vu-Mid/fbvdv N/mm2







The Nominal Shear Resistance Vn at any Section of Girder is the Lesser value Computed by the Equations

i) Vn = Vc + Vs + Vp (Equ. 5.8.3.3-1) &












Computation of Value b & q at different Locations of Girder as per AASHTO-LRFD-5.8.3.4:

Calculation of Longitudinal Strain in Web Reinforcement εs in mm/mm on the Flexural Tension side of Girder withEquations ;




d)

e) 0.26

f) -

g)

1,256.637

1,256.637

h) - MPa.multiplied by Locked-in differencein Strain between the Prestressing Tendonsand Surrounding Concrete (Mpa). For the usal level of Prestressing, the value


j)

k)

I)

Location Length of Factored Factored Calculated Calculated Effective

of Section Segment Moment Shear value of value of Moment

from from the for the for the Effective value

Support Earlier Section. Section. Shear Depth

Section

mm kN-m kN mm kN kN-m

At Faces 1.650 72.156 268.351 1647.000 441.974 441.974




Ac is Area of Concerte on Flexural Tention side of Girder in mm2 having value Ac mm2

Ac = bX-Web* hX-Gir,/2 mm2



As is Area of Non-Prestressing Steel on Flexural Tention side of Girder for the


i) On Faces of Main Girder Position value of Provided Steel Area in mm2 As-Faces. mm2

ii) At Midle of Cross-Girder the Provided Steel Area in mm2 As-Mid. mm2






Mu is Factored (+) ve Moment quantity of the Section in N-mm but not less then Vudv.




Vudv

MuEff



At Middle 0.825 174.887 168.744 1647.000 277.921 277.921

vi)

a) 0.034

b) 0.022

vii)provisions of Equation 5.8.3.4.2-1 :

a)

b) 1.746

c)

d) 1.000

d) 0.629

j)

viii)

a)

Location of Section From Referece Value of Support of Table q b

a) At Faces of Main Girder 5.8.3.4.2.-1 0.034 1.000 36.40 2.23b) At Middle if Cross-Girder Span 5.8.3.4.2.-1 0.022 0.629 33.70 2.38

b)

1.356

1.499

ix)


Value of vc/f/c at Face Pisition of Main Girder = vc-Face/f/

c vc-Face/f/c

Value of vc/f/c at Middle of Cross-Girder = vcMid./f/

c vcMid./f/c


Since for RCC Girder values of Prestressing Components Nu, Vp, Asp, fpo, Ep etc. are = 0, thus equation



Considering the Initial value of ex = 0.001at Support Position & the Effective cotq

Moment Mu-Eff. From the Equation Cotq = (ex2EsAs - Mu/dv)/0.5Vu


At Faces of Main Girder Position = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Face*1000

At Middle of Cross-Girder Span = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-MId*1000

Since both ae Faces & Mid of Cross-Girder Span values of es-x1000 > 0.75 ≤ 1.00 , thus values of q & b for these

Sections can obtain from Table -5.8.3.4.2-1 in respect of values of vc/f/c.

Computation of Values of q & b from AASHTO-LRFD'sTable -5.8.3.4.2-1. against the Respective Calculated

values of esx1000, Ratio vc/f/c :


vc/f/c exx1000


i) At Faces of Maingirder value of Cotq Cotq

ii) At Middle of Cross-Girder Span value of Cotq Cotq



a)


x) Regions Requiring Transverse or Shear/Web Reinforcements under AASHTO-LRFD-5.8.2.4 :


b)


d) - N

e)

f) f 0.90

g)

h)

Location from Girder's Values of Relation Equation

Bearing Center. b Between Satisfied/

N N N Not Satisfied

a) At Support (Bearing Center) Position 2.230 268350.80 33,608.97 15124.035 Vu>0.5Vc Satisfied

a) At a Distance 0.150m from Support 2.380 168743.750 5,739.15 2582.615 Vu>0.5Vc Satisfied

i)

xi) Checking of Required Max. Spacing for Transvers/Shear Reinforcement due to Applied Shearing Stress on Girder under provision of AASHTO-LRFD-5.8.2.7 :

a)

b)

Since the Critical Section is at the Faces of Main Girder, thus Values of q & b of that Section are Governing Values

for Computation of Nominal Shearing Strength of Concrete (Vc) for the Girder.








f is Resistance Factor according to AASHTO-LRFD-5.5.4.2. having value 0.90



Vu Vc 0.5fVc

Vu & Vc

The Table indicates that all the Sections of RCC Girder have satisfied the required provisions for Transverse/Shear Reinforcements under Equation Vu > 0.5f (Vc + Vp). AASHTO-LRFD-Equation-5.8.2.4-1.







d)


Segment. Value of Value of Value of Relation Value of Value of Relation Relation Formula

Location between between between Status

(Between for the for the for the Whether

2-Section) Section Maximum Section Section Satisfy or

mm mm mm Not Satisfy

Section at 1647.00 0.724 2.625 vu<0.1.25f/c 1317.600 658.800 0.8dv>600 Not Satisfy

Faces 0.4dv>300 Not Satisfy


Middle 0.4dv>300 Not Satisfy

e) The Table indicates that non of the Sections of Cross-Girder have satisfied the required provisions for Spacings of the

xii) Chacking for Transverse/Shear/Web Reinforcements as Deep Beam Component (AASHTO-LRFD-5.8.1.1) :

a)Support is Considered as a Deep Component. Deep Beam Components the Shear Reinforcements are being Provide

b) 268.3508 kN. 268.351 kN

c) 134.18 kN 134.175 kN

d) 168.744 kN. 168.744 kN

e) Vu-Mid >1/2Vu-Face.

f)Span is greater than 1/2Vu-Face, thus the Cross-Girder can consider as a Deep Beam Component. Accordingly

xiii) Detaling of Requirments for Deep Beam Component to Provide Transverse/Shear/Web Reinforcements for Girder under provision of AASHTO-LRFD-5.13.2.3 :

a) To provide Transverse/Shear/Web Reinforcements at Different Section of Girder, it should Satisfy the Equation,

b) N


c N/mm2


dv vu 0.125f/c 0.8dv 0.4dv

vu & 0.125f/c 0.8dv & 0.4dv &

sMax-1<600 sMax-2<300


Transverse/Shear Reinforcement under Equations, For vu < 0.125f/c ; smax. = 0.8dv ≤ 600mm, (AASHTO-LRFD-Equ.

5.8.2.7-1). & For vu > 0.125f/c ; smax. = 0.4dv ≤ 300mm, (AASHTO-LRFD-Equ.5.8.2.7-2).

The Component in which a Load causing more than 1/2 of the Shear at a Distance closer than 2d from the Face of

according to Provisions of AASHTO-LRFD-5.6.3 (Provisions of Strut-and-Tied Model) & AASHTO-LRFD-5.13.2.3.

Factored Shear Force at Face of Main Girder, Vu-Face.= Vu-Face.

Value of 1/2 of Factored Shear Force at Face = 1/2Vu-Face.

Factored Shear Force at Middle of Cros-Girder Vu-Mid = Vu-Mid

Status in between the values of 1/2Vu-Supp. & Vu-Mid

Since the Distance 2d from Face of Main Girder is beyond the Cross-Girder Span & value of Shearing Force at Mid

its Transverse/Shear Reinforcementsb can provide under Article 5.13.2.3. Of AASHTO-LRFD.

Ng = f fyAs ³ 0.83bvs (AASHTO-LRFD-5.13.2.3). Here,



c) 250 mm

d) 410.00 MPa

e)

f) 300 mm

g) 300 mm

xxi) Computation of Spacing for Transverse/Shear Reinforcement at Different Section of Girder :

a) d/4 475 mm

b) Allowable Max. Spacing for Transverse/Shear Reinforcement 300 mm

c) 10 mmof Vertical Stirrups.

d) 157.080

d) Let Provide the 175mm Spacing for Transvers/Shear Reinforcements for total 175 mm

xxi) Checking for Requirements of Minimum Transverse Reinforcements for Cross-Girder:

a)

b) Table for Minimum Transverse/Shear Reinforcement at Different Section of Girder.

Location from Bearing Center of Length of s-Web-Bar Status of

Girder. Segment Width Spacing Required Provided Provided &

mm mm mm

Total Length of Cross-Girder. 1650.000 250.000 175.000 40.587 157.080 Av-pro> Av

xxii)

a)



As is Steel Area of Transverse/Shear Reinforcement in mm2 having Spacing s . AS mm2

s is Spacing of Transverse /Shear Reinforcement in mm. The value of s should s-Max

not excide eithe of d/4 or 300mm

The Vertical Spacing, sVetrt. for Crack Control Longitudinal Reinforcement on sVert.-Max

both Vertical Faces of Girder should not Excide either d/3 or 300 mm.

Value of d/4 for the Section under cosideration having d = hGir. (Girder Depth).

s-Max



= 2*pDStir.2/4 mm2

s-X-Gir.

Length of Cross-Girder.



bv' Web Min. Av Av-pro

mm2 mm2 Required Av




Vs = Avfydvcotq /s,


b) 822.122 kN 822.122*10^3 N

c) 908.840 kN 913.806*10^3 N

xxiii)

a)

b)

c) 855.731 kN 855.731*10^3 N

d) 914.579 kN 914.579*10^3 N

xxiv)

a)

b)

c) 2,161.688 kN 2161.688*10^3 N

d) 2,161.688 kN 2161.688*10^3 N

xxv)Computed accordting Provisions of AASHTO-LRFD-5.8.3.3 & AASHTO-LRFD-5.8.2.1 :

a)below ;

b)

c)

d)

e)

At Faces of Main Girder Vs= Avfydv-Face.Cotq/sX-Gir. VS-Face

At Middle of Cross-Girder Span Vs=AvfydvMidCotq/sX-Gir. VSMid.


of AASHTO-LRFD-5.8.3.3 against Equation Vn = Vc + Vs + Vp (Equ. 5.8.3.3-2) :



At Faces of Main Girder Vn= Vc-Face + Vs-Face. Vn-Vace-1

At Middle of Cross-Girder Vn = Vc-Mid. + Vs-Mid. Vn-Mid-1


of AASHTO-LRFD-5.8.3.3 against Equation Vn = 0.25f/cbvdv + Vp (Equ. 5.8.3.3-2) :



At Faces of Main Girderr Vn= 0.25f/c-bv-Facedv-Face. Vn-Face.-2

At Middle of Cross-Girder Vn= 0.25f/c-bvMiddv-Mid. Vn-Mid-2

Accepted Nominal Shear Resistance-Vn & Factored Shearing Resistance-Vr at Different Section of Girder




For RCC Girder, Vp = 0.



f) N

g) 0.90

h)

i)

Section Calculated Relation Accepted Factored Relation Status

Location Factored As per As per Equ. between Value of Shear between If Vr > Vu, the

from Center Shear Equation. 5.8.3.3-2 Values of Resitance Values of Structure is

of Bearing. 5.8.3.3-1 Safe otherwise

kN kN kN kN kN No Safe.

At Faces 268.351 855.731 2161.688 Vn-1< Vn-2 855.731 770.16 Vr> Vu Structure Safe

At Mid Span 168.744 914.579 2161.688 Vn-1< Vn-2 914.579 823.12 Vr> Vu Structure Safe

xxx)of Girder, Computed under Provisions of AASHTO-LRFD-5.8.3.3 :

a)are shown in the Table blow :

Location of Section from Bearing Calculated Computed Status Center of Girder. value of value of between

Values of

kN kNa) At Faces of Main Girder 268.351 855.731 Vu< Vn

b) At Middle of Cross-Girder 168.744 914.579 Vu< Vn

b)is Safe in respect of Applied Shearing Forces caused by Dead Load & Live Loads to the Bridge Structure.

xxxi)

a) Against Applied Loads (DL & LL) the Critical Section for Shearing Forces Prevails at the Face of Support, which is

b) Calculated Nominal Shearing Resistance for Girder on Face of Main Girder, 855.731 kN

855.731*10^3 N

c) 250 mm




The Acceptable Nominal Shear Resistance, Vn, Respective Factored Shear Resitance-Vr at Different Section of

T-Girder, the Staus between the vaues of Vn Computed under Equ. 5.8.3.3-1 & Equ. 5.8.3.3-2 are shown in Table below :.

Table-; Accepted Nominal Shear Resistance-Vn & Factored Shear Resitance-Vr at Different Section :

Vn-1 Vn-2

Vn

Force-Vu Vn-1 & Vn-2 Vr Vr& VU

Vn-1 > Vn-2 Vr> VU

Relation between Factored Shearing Force (Vu) & the Nominal Shear Resistance (Vn) at Different Section

The Relation between Factored Shearing Force, Vu & the Nominal Shear Resistance Vn at Different Section of Girder

Table-; Showing between Factored Shearing Force, Vu & the Nominal Shear Resistance Vn :

Vu Vn

Vu & Vn

Since the Factored Shearing Forces Vu < Vn < Vr, the Computed Nominal Shear Resistance, thus the Girder

Checking of T-Girder Web Width (bWeb) in respect of Nominal Shearing Resistance (Vn) at Critical Section :

at a Distance 0.150m from Girder's Bearing Center (AASHTO-LRFD-5.8.3.2).

Vn

Vn = 860.408*10^3 N.

Provided Girder Width (Web Width of T-Girder) = bX-Web mm. bX-Web


e) 105.771 mm

for the Critical Section.

f)

xxxii) Checking in respect of Longitudinal Reinforcements (Tensile Reinforcements) Provided for Girder under Provision of AASHTO-LRFD-5.8.3.5 :


Where;

b) Variable

c) 410.00 MPa

d) -

e) - MPa

f) Variable N-mm


h) Variable mm

i) Variable N

j) Variable N

k) - N


m) 0.90 AASHTO-LRFD-5.5.4.2.

In RCC Girder the value of Vp = 0, thus according to Equ. 5.8.3.3-2, the Width bv-Cal.

of cross-Girder, bv = Vn/0.25*f/cdv, where dv is Calculated Effective Shear Depth

Since the Calculated value of Girder width bv-Cal = bv the Provided Girder Width, thus Design Critical Section is OK.




















n) 0.90 AASHTO-LRFD-5.5.4.2,

o) 1.00 AASHTO-LRFD-5.5.4.2.

p)

q) Table- Showing Evalution of Equation-5.8.3.5-1 at Different Section of Girder & Status of Results.

Location of Section Calculted Calculted Status

from Bearing Center Provided Factored Factored Shearing Value of R/H Part of the

of Girder. Tensile Moment Shearing Resistance of the EquationSteel Area Force of Stirrups (L/H Part) Equation Satisfied

kN-mm kN kN kN kN Not Satisfieda)At Faces 1256.637 441.974 268.351 822.12243618 515.221 88.390 Satisfiedb) At Middle 1256.637 277.921 168.744 908.84000113 515.221 (248.371) Satisfied

j) Since at all Section the requirments of Equation are being Satisfied, thus provision of Transverse/Shear Reinforcements for Girder is OK.

xxxiii) Checking for Factored Tensile Resistance of Transverse/Shear Reinforcements Provided for Girder:

a) The provided Transverse/Shear/Web Reinforcements at Different Section of Girder under the Provision of Deep Beam

b) Variable N

c) 250 mm

d) 410.00 Mpa

e) 157.080

f) s Variable mm

g) The Girders are being provided with Transverse/Shear Reinforcements in the form of Vertical Stirrups against Shear Forces caused by the Dead Load & Live Loads applied to the Bridge Structure. Due to Vertical Positioning of those Transverse/Shear Reinforcements they will also under Tensile Stress caused by the same Shearing Forces whose actions are in some extent in Vertical Direction. On these back drop mentioned Shearing Forces at different Sectionof Girder may be considered as Tensile Forces for Vertical Stirrups. Thus the Factored Shearing Forces at Each





As. Mu Vu Vs

Asfy

mm2

should Satisfy the Equation, Ng = f fyAs ³ 0.83bvs (AASHTO-LRFD-5.13.2.3; Equ-5.13.2.3-1). Here,




As is Steel Area of Transverse/Shear Reinforcement in mm2 having the Spacing AS mm2

s mm. Here As = Av, the X-Sectional Area of 2-Leged the 10f Dia. Vertical Stirrups.

s is Spacing of Transverse /Shear Reinforcement in mm. The value s should not Excide eithe d/4 or 300mm


h) The Checking for Factored Tensile Resistance of Transverse/Shear Reinforcements at Different Section of Girder areshown in the under mention Table:

i) Table- Showing Minimum Transverse/Shear Reinforcement at Different Section of Cross-Girder.

Location of Section from Bearing s'-Spacing Calculted Calculted Status of

Center of Girder. of Shear Provided Value of Value of Equation

Bars Steel Area Satisfiedmm Not Satisfied

a) At Faces of main Girder 175.000 157.080 57962.384 36312.500 Satisfiedb) Middle of Cross-Girder 175.000 157.080 57962.384 36312.500 Satisfied

j)thus Provision of Transverse/Shear Reinforcements for Cross-Girders are OK.

11 Design of Bridge Girder & Cross-Girder against Earthquake Forces including Checking:

i) According to AASHTO-LRFD-4.7.4.2 for Single-Span Bridges no Seismic analysis is required regardless ofLocation Seismic Zone.

ii) In Single-Span Bridges the Connections between Superstructure & Substructure on Abutment Cap/Seat& Bridge Bearings should be Designed according to provisions of AASHTO-LRFD-3.10.9 & 4.7.4.4.

iii) Design of Abutment Cap/Seat & Bridge Bearings are being done in Respective Design Sheets.

Section, Vu = Ng, the Factored Tensile Resistance Carried by the Pair of Transverse Reinforcement.

As

ffyAs 0.83bvsmm2

Since in all Section the requirments of Equation Equ-5.13.2.3-1, Ng = ffyAs > 0.83bvs are being Satisfyed,


Mr)


Not Satisfy

Not Satisfy

Not Satisfy

Not Satisfy

Not Satisfy

Not Satisfy

Girder have satisfied the required provisions for Transverse/Shear


Computed under Equ. 5.8.3.3-1 & Equ. 5.8.3.3-2 are shown in Table below :.


Mr)


Not Satisfy

Not Satisfy

Not Satisfy


Computed under Equ. 5.8.3.3-1 & Equ. 5.8.3.3-2 are shown in Table below :.

I. Service Limit State Design (WSD) of Main Girder & Cross Girders Against Applied Forces :

1 Data for Flexural Design :


i) Dimensional Data of Superstructure :




d) Thickness of Deck Slab 0.20 m

e) Thickness of Wearing Course 0.08 m

f) Number of Main Girders 5 nos

g) Number of Cross Girders 5 nos

h) Depth of Main Girders (Including Slab as T-Girder) 2.00 m

i) Depth of Cross Girders (Excluding Slab) 1.70 m

j) Width of Main Girders 0.35 m

k) Width of Cross Girders 0.25 m

l) C/C Distance Between Main Girders 2.00 m

m) Distance of Slab Outer Edge to Exterior Girder Center 1.13 m

n) Clear Distance Between Main Interior Girders 1.65 m

o) Filets : i) Main Girder in Vertical Direction 0.15 m

p) ii) Main Girder in Horizontal Direction 0.15 m



ii) Design Data related to Live Loading:

a) Design Criterion : AASHTO Load Resistance Factor Design (LRFD).


iii) Design AASHTO HS20 Truck Loading :









iv) Design AASHTO Lane Loading :

SL

SAddl.

LGir.

hSlab.

hWC

NGir.

NX-Gir.

hGir.

hX-Gir.

bGir.

bX-Gir.

C/CD-Gir.

CD-Ext.-Gir-Edg.

ClD-Int.-Gir.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

DAxel.

DWheel.

LLRW-Load

LLRS-Load

LLMW-Load

LLMS-Load

LLFW-Load

LLFS-Load

a) Design Lane Loading is an Uniformly Distributed Load having Magnitude of 9.30 N/mm 9.300N/mm through the Length of Gridge for Single and acting over a 3.000m 9.30 kN/m Wide Strip in Transverse Direction.

v) Design AASHTO Pedestrian Loading :



vi)

9.807

Unit weight of Normal Concrete 2,447.23

Unit weight of Wearing Course 2,345.26

Unit weight of Normal Water 1,019.68

Unit weight of Saline Water 1,045.17

Unit weight of Earth (Compected Clay/Sand/Silt) 1,835.42

vii)







a) 21.00 MPa

b) 8.40 MPa

c) 23,855.62 MPa

c) 2.89

d) 2.89 MPa

e) 410.00 MPa

f) 200000 MPa

ix) Strength Data related to Working Stress Design (WSD) under Service Limit State ( SLS ) According toAASHTO-LRFD-5.7.1:

a) 8.40 MPa

b) 164.00 MPa

LL-Lane

LL-Pedest N/mm2




gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wWater-Nor. kN/m3

wWater-Sali. kN/m3

wEatrh kN/m3


c



c Ec










c) n 8

d) r 20e) k 0.30 f) j 0.90

g) R 1.14

x) Design Data for Site Conditions :


b) Velocity of Wind Load in Cyclonic Storm Condition 260.00 km/hr


2 Design Phenomena & Calculations for Monent & Shear :

i) Design Phenomena under Service Limit State (WSD) :

a)

b)

respective Moments & Shears. Since the Bridge Deck Slab is integral Part of Girders, thus the Design of Girders

ii) Calculation of Flange Width for Girders :

a) Cross Sectional Sketch Diagram of Bridge Girders & Dack Slab :1.475

0.23

7.30 1.25

0.30

1.070

0.300

0.200

0.25

0.950 1.65 1.650 1.65 1.650 0.95

1.13 2.00 2.000 2.00 2.000 1.125

10.25

b)

= 6.10 m

Modular Ratio, n = Es/Ec ³ 6



VWL-Nor.

VWL-Spe.

VWA

The Flexural of Girders will be according to AASHTO LRFD or Ultimate Strength Design (USD) Procedures but forChecking in respective Issues, Designing under Service Limit State (WSD) up to Certain Level would be done.

Since the Interior Girders of the Bridge have the Max. Moments & Shearing Forces caused by Applied Loads (DL& LL), thus it is require to conduct the Flexural Design for Reinforcements of Bridge Girders Based on Calculated

will be under T-Beam if the Provisions in these Respect Satisfy, otherwise Designee will be under Provisions for theRectangular Beam.

Calculation of Effective Flange Width of T-Girder under AASHTO LRFD-4.6.2.6 (4.6.2.6.1) as least Dinention of :

* One-quarter of Effective Span Length = 1/4*SL

CLCL

2.75 m

= 2 mSince Average Spacing of Adjacent Beams/Girders is the Least one,

2.00 m

c) From Load, Shear & Moment Calcutation Tables it appares that, the Interior Girders are facing the Max. Resultant

iii) Calculations for Monent at Different Location of Girder :

a)

Table-1. Sum. of Max. Moments Against All Applied Loads (DL & LL) on Interior Girder.


0.000 142.152 1039.048 1811.021 2263.908 2442.182 2449.720

0.000 28.627 207.537 357.399 449.585 483.808 484.096

0.000 79.345 563.836 941.812 1133.930 1201.915 1140.188


b) Moment at Sopport Position of Girder 0 kN-m

c) 119.500 kN-m

d) 1,810.421 kN-m

e) 3,110.232 kN-m

f) 3,847.423 kN-m

g) 4,127.905 kN-m

h) 4,074.004 kN-m

i) Sketch Diagram of T-Beam/Girder :

2.000

0.200

1.826 2.000

* 12.0 times average Depth of Slab + Greater Thickness of Web = 12*hSlab + WGir. = * One-half the Width of Girder Top Flange (It is not req. as there is no Addl. Top Flange)

* The average Spacing of Adjacent Beams/Girders = C/CD-Gir.

thus the Flange Width of Interior Girders, WFlang = 2.000m bFlan.

Forces (DL & LL) causing Max. Shears & Moments, thus One of Interior Girders is considered for as Typical one for Flexural Design in respect of All Applied Loads (DL + LL) and Corresponding Moments & Shears.

Table for Max. Moments at Different Locations of Interior Girder due to Factored DL, Lane-LL & Wheel-LL :



c. Wheel Live Load (WLLInt)

Mu-Support.

Moment at a Distance 0.150m from Sopport of Girder Mu-0.150m.



Moment at a Distance 3L/8 from Sopport of Girder Mu-3L/4.

Moment at Absolute Max. Moment Loaction (c.g. Position) of Girder Mu-c.g.

Moment at a Distance L/2 (Middle of Span) from Sopport of Girder Mu-L/2.

bFln.

hFln. =

d = hGir. =

0.350

iv) Factored Flexural Resistance for Prestressed or RCC Structural Components (AASHTO-LRFD-5.7.3.2.1):

a) N-mm

N-mm

f

b)

c) For a Nonprestressing Structural Component of Rectangular Elements having Singly Reinforced, , at any Section the

v) Related Features for Working Stressed Design of Main Girder with Max, Moment value (At c.g. Position) :

a) 4,127.905 kN-M

4,127.905 kN-m, thus 4127.905*10^6 N-mmthis value is Considerd as the Moment at Middle Position of Span.

b) 50 mm

50 mm

38 mm

c) 32 mm

d) 804.25

e) 32 mm

f) 12 mm

g) Thus Effective Depth of Reinforcements from Top of Girder up to Center of the 1,826.00 mm

h) b 2000 mm

i) 200 mm

j) 0.008

vi) Checking's Whether the Bridge Girder would Designed as T-Beam or Rectangular Beam Provisions :

bWeb =




For a Nonprestressing Structural Component either of I or T Section having Flenge & Web Elements, at any Section

the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2)

Nominal Resistance, Mn = Asfy(de-a/2)

The Absolute Max. Moments on Interior Girder is at c.g. Point. Since it is very MMax.

close to Middle Position of Span having value MMex. =

Let the Clear Cover at Bottom Surface of Girder, C-Cov.Bot. = 50mm, C-Cov-Bot.

Let the Clear Cover at Top of Girder, C-Cov.Top = 50mm, C-Cov-Top.

Let the Clear Cover at Vertical Faces of Girder, C-Cov.Vert.. =38mm, C-Cov-Side.

Let the Main Reinforcements are 32f Bars in 4 Layers, DBar


The Vertical Spacing between Reinforcement Bars, SVer. = 32 mm SVer.

Let the Transverse/Shear Reinforcements (Stirrups) are of 12f Bars, DStir.

dL/2

Provided Reinforcements d = (hGir - C-Cov-Bot - DStri -2*DBar- 1.50*SVer.)

Flange Width of T-Girder, b = 2.000 m

Thickness of T-Girder Flange, hf = 200mm hf

According to WSD Method the Balanced-Stress Steel Ratio for Girder Section pe.

pe = n/2r(n + r)

a) According to Working Stressed Design Provisions a Rectangular having Flange with Reasonable Thickness on its

b)Rectangular Beam.

c) kd 548.552 mm hf<kd T-Beam Provision

d)

vii) Design of Tensile Reinforcements for Intermediate Girders against Calculated Moments :

a) 14,582.938

b) 18.132 nos.

c) 18.000 nos.

d) 1,808.222 mm

Top should be Designed as T-Beam if Depth of Netural Axis 'kd' is Less than the Flange Thickness.


According to WSD Provisions the depth of Neutral Axis from Compressiof Face is Expressed by kd.

Since the Calculated Depth Neutral Axis kd < hf, Thickness of Flange, thus in Working Stressed Design (WSD) the Provisions for T-Beam will prevail.

Under WSD for T-Girder the Required Flexural Steel Area, As = M/fs(d-hf/2) As-req. mm2

Number of 32f Bars required as Main Reinforcement of T-Girder, NBar-req

NBar-req = As-req./Af-32

Let Provide 18 nos. 32f Bars as Main Reinforcement for the T-Girder on ite NBar-pro Mid Span (L/2) since under USD the Provided number of Bars for the Section is 17 nos. 32f Bars.

Effective Depth of Reinforcements is Same as that of USD Design. de

J.

1

2.147 m

9.350 m

0.450 m 0.450 m

2 Structural Data :

Description Notation. Value. Unit.

i) Dimentions of Abutment Cap & Back Wall :

a) Height of Back Wall 2.147 m

b) Thickness of Back Wall 0.300 m

c) Thickness of Wing Wall 0.450 m

d) Length of Back Wall between the Wing Walls 9.350 m

e) Length of Abutment Cap between outer Faces of Wing Walls 10.250 m

f) Height of Abutment Cap Rectangular Portion (Abutment Seat) 0.600 m

g) Height of Abutment Cap Trapezoid Portion (Abutment Seat) 0.300 m

h) Width of Abutment Cap Rectangular Portion (Abutment Seat) 1.000 m

i) Width of Abutment Cap Trapezoid Portion at Bottom(Abutment Seat) 0.450 m

j) Offset of Trapezoid Portion on Earth Face 0.400 m

k) Offset of Trapezoid Portion on Water Face 0.150 m

3 Design Criterion, Type of Loading & Design Related Data :




ii)

9.807

Structural Design of Back Wall for Abutment :

Sketch Diagram of Back Wall:

hB-Wall =

LB-Wall =

tW-Wall= tW-Wall =

hBack-Wall

tBack-Wall

tWing-Wall

LBack-Wall

LAb-Cap

hAb-Cap-Rec

hAb-Cap-Tro.

bAb-Cap-Rec

bAb-Cap-Tro-Bot.

bOffset-Earth.

bOffset-Water.



35 035 0

30 0

700

60 0

300

2147

4047

450

150400

2147

60 0






iii)







a) 21.000 MPa

b) 8.400 MPa

c) 23855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.00 MPa

g) 164.000 MPa

h) 200000 MPa











j) 0.850 (AASHTO LRFD-5.7.2..2.)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


c



c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).









fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.


k) 0.850

vi) Other Design Related Data :






a)

Here:


1.000

1.000

1.000

ii) Selection of Load Multiplying Fatcors for Strength Limit State Design (USD) & Load Combination :


iii) Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :




VWL-Nor.

VWL-Spe.

VWA










Qi = Force Effect,



but those would be occasional. Thus the respective Multiplier Factors of Limit State STRENGTH I (Bridge used by Normal Vehicle without wind load) for normal operation, Limit State of STRENGTH-III (Wind Velocity exceeding90km/hr) for wind load during cyclonic storm condition and Limit State of STRENGTH-IV (Having Wind Velocity of 90 km/hr) for normal wind load only are selected as CRITICAL conditions for bridge structure.








iv) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000



gEH


gEV


gES


















q) -

r) -

t) 1.000












b) 1.000

c) IM 1.000









gEH


gEV


gES




AASHTO LRFD-3.6.2.1, Table 3.6.2.1-1; SERVICE - I(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

6 Load Coefficients Factors & Intensity of Different Imposed Loads :


a) 0.441



















b) f 34.000



d) 0.34 to 0.45 dim



0.007935


c) 0.018000

18.000




0.004761

4.761



0.00476052

4.761

c) k 0.441

d) 1,835.424

e) g 9.807

f) 900.000 mm

600.000 mmAASHTO-LRFD-3.11.6.4; Table-3.11.6.4-1.

g) Width of Live Load Surcharge Pressure for Abutment having 900.000 mm 0.900 m




O

Value of Tan d (dim) for Coefficient of Friction. tan d = 0.34 to 0.45 (AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.)

Dp-ES kN/m2

Dp-ES = ksqs in Mpa. Where; N/mm2


qs is Uniform Surcharge applied to upper surface of Active Earth Wedge(Mpa) wE*10-3 N/mm2

= wE*10-3N/mm2 kN/m2


kN/m2


kN/m2


kN/m2


kN/m2






Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.


h) 1,050.000 mm


i) Width of Live Load Surcharge Pressure for Wing Walls, 600.000 mm 0.600 m

600.000 mm 0.600 m

7 Philosophy in Flexural Design of Back Walls of Bridge Structure :

a)

Cantilever one. In Construction Phenomenon execution Back Wall works are to be done a well ahead before the

as a Cantilever Flexural Component.

b) 1.000 mForces & Moments for Back Wall it is considered having 1.000m Length & Wide Strips according to sequences of Design.

8

i) Horizontal Loads at Bottom Level of Back Wall on 1 (One) meter Strip :

a) 17.035

b) 7.935

c) 7.141

ii) Factored Horizontal Loads on Back Wall for 1 (One) meter Wide Strip (Strength Limit State-USD):

a) 25.553

b) 11.902

Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

The Back Wall of Bridge Substructure will have to face Horizontal Dead Load (DL) & Live Load (LL) Pressures due to Earth & Surcharge on Back Faces. Under Horizontal Pressure (DL & LL) the Back Wall will behave as a

Construction of Super structural Components having the total Horizontal Loads (DL & LL) upon it. Thus to ensure the sustainability of Back Wall against the affect of Horizontal Loads & Forces (DL & LL) it should be Designed

For the Design purpose & Calculation of Imposed Loads (DL & LL), Shearing LH&V

Computation of Horizontal Loads on Back Wall due to Lateral Soil & Surcharge Pressure :

Horizontal Pressure Intensity due to Lateral Soil Dead Load (DL) at PDL-H-Soil-Bot. kN/m2/m

Bottom Level of Back Wall (Perpendicular to Traffic). = k0*wE*hBack-Wall

Horizontal Pressure Intensity due to Surcharge Dead Load (DL) on PDL-H-Sur kN/m2/m

Back Wall (Perpendicular to Traffic). = Dp-ES.LH&V

Horizontal Pressure Intensity due to Surcharge Live Load (LL) on PLL-H-Sur kN/m2/m

Back Wall (Perpendicular to Traffic) having hBack-Wall < 6.000m. = Dp-LL-Ab.LH&V

Factored Horizontal Pressure Intensity due to Lateral Soil Dead FPDL-H-Soil-Bot.-USD kN/m2/m

Load (DL) at Bottom Level of Back Wall (Perpendicular to Traffic). = gEH*PDL-H-Soil-Bot

Factored Horizontal Pressure Intensity due to Surcharge Dead Load (DL) FPDL-H-Sur-USD kN/m2/m

on Back Wall (Perpendicular to Traffic). = gES*PDL-H-Sur

c) 12.496

d) Total Factored Horizontal Pressure Intensity due to All Applied Loads 49.951

iii) Factored Horizontal Loads on Back Wall for 1 (One) meter Wide Strip (Service Limit State-WSD):

a) 17.035

b) 7.935

c) 7.141


9 Computation of Coefficients for Calculation of Moments on Back Walls :

i) Span Ratio of Back Walls in respect of of Horizontal to Vertical Spans :

a) Horizontal Span Length of Back Wall (Inner Distance) 9.350 m

b) Vertical Height of Back Wall (From Back Wall Top upto Abutment Cap Top) 2.147 m

c) Bach Wall Span Ratio for of Horizontal to Vertical with Unsupported on Top 4.355

d) Bach Wall Span Ratio for of Vertical to Horizontal with Unsupported on Top 0.230

ii) Coefficients for Calculation of Moments related to Triangular Loadings on Back Walls :(Using Table-53 of Reinforced Concrete Design Handbook UK.)

a) k 3.000

b) (+) ve Moment Coefficient of Back Wall for Vertical Span. 0.000

c) (+) ve Moment Coefficient of Back Wall for Horizontal Span Strip. 0.000

d) (-) ve Moment Coefficient of Back Wall for Vertical Span Strip on Bottom 0.003 Surface at Middle Position under Triangular Loading at Bottom

Factored Horizontal Pressure Intensity due to Surcharge Live Load (LL) FPLL-H-Sur-USD kN/m2/m

on Back Wall (Perpendicular to Traffic). = mgLL-LS*PLL-H-Sur-Ab

åFPH-B-Wall-USD kN/m2/m(DL & LL) at Bottom Level of Abutment (Perpendicular to Traffic).

Factored Horizontal Pressure Intensity due to Lateral Soil Dead FPDL-H-Soil-Bot-WSD kN/m2/m

Load (DL) at Bottom Level of Back Wall (Perpendicular to Traffic). = gEH*PDL-H-Soil-Bot

Factored Horizontal Pressure Intensity due to Surcharge Dead Load (DL) FPDL-H-Sur-WSD kN/m2/m

on Back Wall (Perpendicular to Traffic). = gES*PDL-H-Sur

Factored Horizontal Pressure Intensity due to Surcharge Live Load (LL) FPLL-H-Sur-WSD kN/m2/m

on Back Wall (Perpendicular to Traffic). = mgLL-LS*PLL-H-Sur-Ab

åFPH-B-Wall-WSD kN/m2/m(DL & LL) at Bottom Level of Abutment (Perpendicular to Traffic).

LBack-Wall

hBacl-Wall

kUnsupp.-H/V

= LBack-Wall/hBack-Wall

kUnsupp.-V/H

= hBack-Wall/LBack-Wall

Since k = 4.355 > 3.00, thus let k = 3.000

(+)cS-V-BW-T

(+)cS-H-BW-T

(-)cS-V-BW-T

e) (-) ve Moment Coefficient of Back Wall for Horizontal Span Strip on Faces 0.000of Sides.

iii) Coefficients for Calculation of Moments related to Uniformly Loadings on Back Wall :(Using Table-4.2; 4.3 & 4.4 of Book Design of Concrete Structures- G.Winter)

a) k 0.500

a) 0.056 Span Strip.

b) 0.004 Span Strip.

c) 0.076 Strip.

d) 0.005 Span Strip.

e) 0.089 Back Wall Vertical Span Strip

f) 0.010 Back Wall Horizontal Span Strip

10 Factored Moments due to Horizontal Loads on Back Walls Under Strength Limit State (USD):

i)

a) (+) ve Moment on Vertical Span due to Traingular Soil Load - kN-m/m

b) (+) ve Moment on Vertical Span due to Uniform Surcharge Dead 3.072 kN-m/m

c) (+) ve Moment on Vertical Span due to Uniform Surcharge Live Load 4.378 kN-m/m

d) Total (+) ve Moment on Vertical Span Strip for all Applied Loads 7.450 kN-m/m

ii)

a) (-) ve Moment on Vertical Span due to Traingular Soil Load 0.294 kN-m/m

b) (-) ve Moment on Vertical Span due to Uniform Surcharge Dead Load 4.883 kN-m/m

(-)cS-H-WW-T

Since k = 0.230 , thus let Consider k = 0.500

(+) ve Moment Coefficient for Dead Load (DL) on Back Wall Vertical (+)cS-V-BW-U-DL

(+) ve Moment Coefficient for Dead Load (DL) on Back Wall Horizontal (+)cS-H-BW-U-DL

(+) ve Moment Coefficient for Live Load (LL) on Back Wall Vertical Span (+)cS-V-BW-U-LL

(+) ve Moment Coefficient for Live Load (LL) on Back Wall Horizontal (+)cS-H-BW-U-LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-V-BW-U-DL+LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-H-BW-U-DL+LL

Calculation of Factored (+) ve Moments on Back Wall in Vertical Span using Coefficients :

(+)MS-V-BW-T-USD

= (+)cS-V-BW-T*FPDL-H-Soil-Bot-USD*hBack-Wall2

(+)MS-V-BW-Sur-DL-USD

Load = (+)cS-V-BW-U-DL*FPDL-H-Sur-USD*hBack-Wall2

(+)MS-V-BW-Sur-LL-USD

= (+)cS-V-BW-U-LL*FPLL-H-Sur-USD*hBack-Wall2

(+)åMS-V-BW-USD

Calculation of Factored (-) ve Moments on Back Wall in Vertical Span using Coefficients :

(-)MS-V-BW-T-USD

= (-)cS-V-BW-T*FPDL-H-Soil-Bot.-USD*hBack-Wall2

(-)MS-V-BW-Sur-DL-USD

c) (-) ve Moment on Vertical Span due to Uniform Surcharge Live Load 5.127 kN-m/m

d) Total (-) ve Moment on Vertical Span Strip for all Applied Loads 10.304 kN-m/m

iii)

a) (+) ve Moment on Horizontal Span due to Traingular Soil Load - kN-m/m

b) (+) ve Moment on Horizontal Span due to Uniform Surcharge Dead 4.162 kN-m/m

c) (+) ve Moment on Horizontal Span due to Uniform Surcharge Live 4.370 kN-m/m

d) Total (+) ve Moment on Horizontal Span Strip for all Applied Loads 8.532 kN-m/m

iv)

a) (-) ve Moment on Horizontal Span due to Traingular Soil Load - kN-m/m

b) (-) ve Moment on Horizontal Span due to Uniform Surcharge Dead 10.405 kN-m/m

c) (-) ve Moment on Horizontal Span due to Uniform Surcharge Live Load 10.925 kN-m/m

d) Total (-) ve Moment on Horizontal Span Strip for all Applied Loads 21.329 kN-m/m

11 Factored Moments due to Horizontal Loads on Back Walls Under Service Limit State (WSD):

i)



c) (+) ve Moment on Vertical Span due to Uniform Surcharge Live Load 2.502 kN-m/m

= (-)cS-V-BW-U-DL+LL*FPDL-H-Sur-USD*hBack-Wall2

(-)MS-V-BW-Sur-LL-USD

= (-)cS-V-BW-U-DL+-LL*FPLL-H-Sur-USD*hBack-Wall2

(-)åMS-V-BW-USD

Calculation of Factored (+) ve Moments on Back Wall in Horizontal Span using Coefficients :

(+)MS-H-BW-T-USD

= (+)cS-H-BW-T*FPDL-H-Soil-Bot-USD*LBack-Wall2

(+)MS-H-BW-Sur-DL-USD

Load = (+)cS-H-BW-U-DL*FPDL-H-Sur-USD*LBack-Wall2

(+)MS-H-BW-Sur-LL-USD

Load = (+)cS-H-BW-U-LL*FPLL-H-Sur-USD*LBack-Wall2

(+)åMS-H-BW-USD

Calculation of Factored (-) ve Moments on Back Wall in Horizontal Span using Coefficients :

(-)MS-H-BW-T-USD

= (-)cS-H-BW-T*FPDL-H-Soil-Bot-USD*LBack-Wall2

(-)MS-H-BW-Sur-DL-USD

Load = (-)cS-H-BW-U-DL+LL*FPDL-H-Sur-USD*LBack-Wall2

(-)MS-H-BW-Sur-LL-USD

= (-)cS-H-BW-U-DL-LL*FPLL-H-Sur-USD*LBack-Wall2

(-)åMS-H-BW-USD

Calculation of Factored (+) ve Moments on Back Wall in Vertical Span using Coefficients :

(+)MS-V-BW-T-WSD

= (+)cS-V-BW-T*FPDL-H-Soil-Bot-WSD*hBack-Wall2

(+)MS-V-BW-Sur-DL-WSD

Load = (+)cS-V-BW-U-DL*FPDL-H-Sur-WSD*hBack-Wall2

(+)MS-V-BW-Sur-LL-WSD

= (+)cS-V-BW-U-LL*FPLL-H-Sur-WSD*hBack-Wall2


ii)





iii)





iv)



c) (-) ve Moment on Horizontal Span due to Uniform Surcharge Live 6.243 kN-m/m


11 Calculation of Unfactored Moments due to Horizontal Dead Loads on Back Walls :

i)


(+)åMS-V-BW-WSD

Calculation of Factored (-) ve Moments on Back Wall in Vertical Span using Coefficients :

(-)MS-V-BW-T-WSD

= (-)cS-V-BW-T*FPDL-H-Soil-Bot.-WSD*hBack-Wall2

(-)MS-V-BW-Sur-DL-WSD

= (-)cS-V-BW-U-DL+LL*FPDL-H-Sur-WSD*hBack-Wall2

(-)MS-V-BW-Sur-LL-WSD

= (-)cS-V-BW-U-DL+-LL*FPLL-H-Sur-WSD*hBack-Wall2

(-)åMS-V-BW-WSD

Calculation of Factored (+) ve Moments on Back Wall in Horizontal Span using Coefficients :

(+)MS-H-BW-T-WSD

= (+)cS-H-BW-T*FPDL-H-Soil-Bot-WSD*LBack-Wall2

(+)MS-H-BW-Sur-DL-WSD

Load = (+)cS-H-BW-U-DL*FPDL-H-Sur-WSD*LBack-Wall2

(+)MS-H-BW-Sur-LL-WSD

Load = (+)cS-H-BW-U-LL*FPLL-H-Sur-WSD*LBack-Wall2

(+)åMS-H-BW-WSD

Calculation of Factored (-) ve Moments on Back Wall in Horizontal Span using Coefficients :

(-)MS-H-BW-T-WSD

= (-)cS-H-BW-T*FPDL-H-Soil-Bot-WSD*LBack-Wall2

(-)MS-H-BW-Sur-DL-WSD

Load = (-)cS-H-BW-U-DL+LL*FPDL-H-Sur-USD*LBack-Wall2

(-)MS-H-BW-Sur-LL-WSD

Load = (-)cS-H-BW-U-DL-LL*FPLL-H-Sur-WSD*LBack-Wall2

(-)åMS-H-BW-WSD

Calculation of (+) ve Moments on Back Wall in Vertical Span due to Unfactored Dead Loads :

(+)MS-V-BW-T-UF


c) Total (+) ve Moment on Vertical Span Strip for all Applied Dead Loads 2.048 kN-m/m

ii)



c) Total (-) ve Moment on Vertical Span Strip for all Applied Dead Loads 3.451 kN-m/m

iii)



c) Total (+) ve Moment on Horizontal Span Strip for all Applied Dead Loads 2.775 kN-m/m

iv)



c) Total (-) ve Moment on Horizontal Span Strip for all Applied Dead Loads 10.405 kN-m/m

12 Computation of Factored Horizontal Shearing Forces on Back Walls :

i) Coefficients for Calculation of Shearing Froces on Back Walls Vertical & Horizontal Faces :(Using Table-4.5 of Book Design of Concrete Structures- G.Winter)

a) k 0.500

b) Coefficiant of Shearing Forces for Vetical Span Strip having Shearing Forces 0.330 Bottom Edge of Back Wall

= (+)cS-V-BW-T*PDL-H-Soil-Bot*hBack-Wall2

(+)MS-V-BW-Sur-DL-UF

Load = (+)cS-V-BW-U-DL*PDL-H-Sur-Bot.*hBack-Wall2

(+)åMS-V-BW-UF

Calculation of (-) ve Moments on Back-Wall in Vertical Span due to Unfactored dead Loads :

(-)MS-V-BW-T-UF

= (-)cS-V-BW-T*PDL-H-Soil-Bot*hBack-Wall2

(-)MS-V-BW-Sur-DL-UF

= (-)cS-V-BW-U-DL+LL*PDL-H-Sur-BW*hBack-Wall2

(-)åMS-V-BW-UF

Calculation of (+) ve Moments on Back-Wall in Horizontal Span due to Unfactored Dead Loads :

(+)MS-H-BW-T-UF

= (+)cS-H-BW-T*PDL-H-Soil-Bot.*LBack-Wall2

(+)MS-H-Ab-Sur-DL-UF

Load = (+)cS-H-BW-U-DL*PDL-H-Sur-BW*LBack-Wall2

(+)åMS-H-BW-UF

Calculation of (-) ve Moments on Back-Wall in Horizontal Span due to Unfactored Dead Loads :

(-)MS-H-WB-T-UF

= (-)cS-H-BW-T*PDL-H-Soil-Bot*LBack-Wall2

(-)MS-H-BW-Sur-DL-UF

Load = (-)cS-H-BW-U-DL+LL*PDL-H-Sur-BW*LBack-Wall2

(-)åMS-H-BW-UF

Since k = 0.230 , thus let Consider k = 0.500

cShear-V-BW

c) Coefficiant of Shearing Forces for Horizontal Span Strip having Shearing 0.670 Forces on Back Wall Vertical Faces

ii) Shearing Froces on Back Walls Vertical & Horizontal Faces Under Strength Limit State (USD):

a) Total Factored Horizontal Load on Back Wall for 1.000m Wide Strip on its 79.814 kN/m

b) Shearing Forces on Back Wall on its Bottom Level for Vertical Span 26.339 kN/m

d) Shearing Forces on Back Wall on its Bottom Level for Horizontal Span 53.475 kN/m

12 Computation of Related Features required for Flexural Design of Vertical & Horizontal Reinforcements of Back Wall :

i) Design Strip Width for Back Wall in Vertical & Horizontal Direction & Clear Cover on different Faces;

a) b 1.000 m = 1000mm

b) 50.000 mm

38.000 mm



b) c Variable

c) Variable

Variable

Variable

410.00

Variable

Variable mm

Variable mm


cShear-H-BW

åFPH&V-BW-USD

Bottom Level = (0.5*FPDL-H-Soil-Bot-USD + FPDL-H-Sur-BW-USD + FPDL-H-Sur-BW-USD)*hBack-Wall*LH&V

VS-V-BW-USD

Strip = cShear-V-BW*åFPH&V-BW-USD

VS-H-BW-USD

Strip on Vertical Faces = cShear-H-BW*åFPH&V-BW-USD

Let Consider the Design Width in both Vertical & Horizontal directions are

Let the Clear Cover on Earth Face of C-Cov.Earth. = 50mm, C-Cov-Earth.

Let the Clear Cover on Open Face of Abutment Wall, C-Cov.Open = 50mm, C-Cov-Open

c/de-Max.











Centroid in mm.








Variable N-mm

Variable

0.015

15.000/10^3

15.000*10^6

2.887

c) 43305340.317 N-mm

43.305 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)




Back 5.7.3.3.2-1 M (1.33*M)Wall kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

thus de = ds .














Snc = (b*tBack-Wall.3/12)/(tBack-Wall./2)










for RCC Mu


2.048 43.305 43.305 43.305 51.966 7.450 9.909 51.966 51.966

Vertical

(-)ve Strip 3.451 43.305 43.305 43.305 51.966 10.304 13.704 51.966 51.966

Vertical

2.775 43.305 43.305 43.305 51.966 8.532 11.347 51.966 51.966

Horizontal

10.405 43.305 43.305 43.305 51.966 21.329 28.368 51.966 51.966

Horizontal

iv)


b) 0.016

13 Flexural Design of Vertical Reinforcements on Earth Face of Back Wall against the (-) ve Moment onVertical Span Strip :


a) The Calculated Flexural (-) ve Moment in Vetrical Span Strip of Back Wall 10.304 kN-m/m

10.304*10^6 N-mm/m

Moment for Provision of Reinforcement against (-) ve Moment value. For (-) ve 51.966 kN-m/mvalue the required Reinforcement will be on Earth Face of Back Wall. 51.966*10^6 N-mm/m

b) 51.966 kN-m/m

51.966*10^6 N-mm/m


a) 12.000 mm

b) 113.097

c) The provided Effective Depth for the Section with Reinforcement on Earth 244.000 mm

d) 12.238 mm

e) 592.021

f) 191.036 mm,C/C

g) 125.000 mm,C/C

(+)ve Strip

(+)ve Strip

(-)ve Strip


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


(-)åMS-V-BW-USD

Wall is Less than the Allowable Minimum Moment Mr. Thus Mr is the Governing

Mr

Since (-)åMS-V-BW-USD> Mr, the Allowable Minimum Moment for the Section, MU


Let provide 12f Bars as Vertical Reinforcement on Earth Face of Back Wall DBW-Earth-V.

X-Sectional of 12f Bars = p*DBW-Earth-V2/4 Af-12. mm2

de-pro.

Face, dpro = (tBack-Wall-CCov-Earth. -DBW-Earth-V./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-BW-Earth-V. mm2/m

Spacing of Reinforcement with 12f bars = Af-12b/As-req-BW-Earth-V. sreq


h) 904.779


a) 0.004

b) 20.782 mm


d) Mpro>Mu OK

e) pmax>ppro OK


a) 0.450

b) c 17.665 mm

c) 0.85

d) 0.072 0.072


v) Checking Against Max. Shear Force on Back Wall in Vetrical Strip at Bottom Level.

a) The Maximum Shear Force occurs at Botton Level of Back Wall on its 26.339 kN/mVertical Strip which is also the Ultimate Shearing Force for the Section. Thus 26.339*10^3 N/m

b)

b-i) 1,000.000 mm


spro = 125mm,C/C

The provided Steel Area with 12f bars having Spacing 150mm,C/C As-pro-BW-Earth-V. mm2/m

= Af-162.b/spro



Mpro


Relation between Provided Resisting Moment Mpro and Calculated Design Moment MU.

Status of Provided Steel Ration rpro in respect of Allowable Max. Steel Ratio rMax.





Thus for the Section the Ratio of c/de = c/de-pro


VU.

Maximum Shear Force, VMax = VS-V-Ab-USD.= VU






219.600 mm 216.000 mm

b-iii) f 0.90

b-iv) - N

c) 709.968 kN/m 709.968*10^3 N/m

709.968 kN/m 709.968*10^3 N/m

1,152.900 kN/m 1152.900*10^3 N/m

c-i) 709.968 kN/m(AASHTO-LRFD- Equ. 5.8.3.3-1); 709.968*10^3 N/m

c-ii) - N/m

c-iii) b 2.000

c-iv) - N

d) Vn>Vu Satisfied




a) 77.993 N-mmwhere; 77.993*10^6 kN-m

86.659 N-mm 86.659*10^6 kN-m

h = tBack-Wall Depth of Well Cap.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-BW-Bot of Equations as mentioned in Aritical 5.8.3.3 :










Relation between Computed Nominal Shear Resitance Vn & Factored Shearing Forces







f 0.90

b)


d) Since Back Wall in Vertical Direction is being considered as a Cantilever 86.659 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 86.659*10^6 N-mm

e) 10.304 kN-m

10.304*10^6 N-mm

f) Mr>Ms-v-bw-usd Satisfied


a)

Where;

b) 28.904

6.381 kN-m 6.381*10^6 N-mm

904.779


c) 249.434





y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-V-BW

at its Bottom will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment M at Bottom of assumed Cantilever Beam (-)åMS-V-BW-USD

in Vertcal on Earth Face = (-)åMS-V-BW-USD


Factored Moment M at Mid Span ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State (-)åMS-V-BW-WSD





i) dc= Depth of Concrete Extreme Tension Face from the Center of the Closest dc


of Rectangular Section, dc = DBar/2 + CCov-Earth. Since Clear Cover on Earth Face of




23,000.000 N/mm


246.000

d)

e) 2,665.210 N/mm


g) fsa > 0.6fy Not Satisfy


i)


j)does require any Longitudinal Skein Reinforcement.

14Vertical Span Strip :


a) The Calculated Flexural (+) ve Moment in Vetrical Span Strip of Back 7.450 kN-m/m

7.450*10^6 N-mm/m

Moment for Provision of Reinforcement against (+)ve Moment value. For (+)ve 51.966 kN-m/m















Since Developed Tensile Stress of Tension Reinforcement of Back Wall fs-Dev.< fs Allowable Tensile Stress; the



More over though the Structure is a Nonprestressed one & value of dc have not Exceeds 900 mm, thus Component

Flexural Design of Vertical Reinforcements on Open Face of Back Wall against the (+) ve Moment on

(+)åMS-V-BW-USD


Mr

value the required Reinforcement will be on Open Face of Back Wall. 51.966*10^6 N-mm/m

b) 51.966 kN-m/m

51.966*10^6 N-mm/m


a) 12 mmWall.

b) 113.097


d) 12.238 mm

e) 592.021

f) 191.036 mm,C/C

g) 125 mm,C/C

h) 904.779


a) 0.004

b) 20.782 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 17.665 mm

Since (+)åMS-V-BW-USD< Mr, the Allowable Minimum Moment for the Section, MU


Let provide 12f Bars as Vertical Reinforcement on Open Face of Back DBW-Open-V.

X-Sectional of 12f Bars = p*DBW-Open-V2/4 Af-12. mm2

de-pro.

Face, dpro = (tBackWall.-CCov-Open. -DBW-Open-V./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-BW-Open-V. mm2/m

Spacing of Reinforcement with 12f bars = Af-12b/As-req-BW-Open-V. sreq


spro = 125mm,C/C

The provided Steel Area with 12f bars having Spacing 150mm,C/C As-pro-BW-Open-V. mm2/m

= Af-12.b/spro



Mpro


Relation between Provided Resisting Moment Mpro and Designed Moment MU.

Relation between Provided Steel Ration rpro and Allowable Max. Steel Ratio rMax.



c) 0.85

d) 0.072 0.072


iv) Since Checking have been done for Provision of Vertical Reinforcement on Earth Face of Back Wall against Max. Moment due to Imposed Loads & found Satisfactory in all respect thus it does not require further Checking against Provision of Vertical Reinforcement on Open Face of Back Wall.

15 Flexural Design of Horizontal Reinforcements on Earth Face of Back Wall against the (-) ve Moment onHorizontal Span Strip :


a) The Calculated Flexural (-) ve Moment in Horizontal Span Strip of Back 21.329 kN-m/m

21.329*10^6 N-mm/m

Moment for Provision of Reinforcement against (-)ve Moment value. For (-)ve 51.966 kN-m/mvalue the required Reinforcement will be on Earth Face of Back Wall. 51.966*10^6 N-mm/m

b) 51.966 kN-m/m

51.966*10^6 N-mm/m


a) 12 mmWall.

b) 113.097


d) 12.908 mm

e) 624.397

f) 181.130 mm,C/C

g) 150 mm,C/C

h) 753.982





(-)åMS-H-BW-USD


Mr

Since (-)åMS-H-BW-USD < Mr, the Allowable Minimum Moment for the Section, MU


Let provide 12f Bars as Horizontal Reinforcement on Earth Face of Back DBW-Earth-H.

X-Sectional of 12f Bars = p*DBW-Earth-H2/4 Af-12. mm2

de-pro.

Face, dpro = (tBack-Wall.-CCov-Earth. - DBW-Earth-V- DBW-Earth-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-BW-Earth-H. mm2/m

Spacing of Reinforcement with 16f bars = Af-12b/As-req-BW-Earth-H. sreq


spro = 150mm,C/C

The provided Steel Area with 12f bars having Spacing 150mm,C/C As-pro-BW-Earth-H. mm2/m


a) 0.003

b) 17.318 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 14.721 mm

c) 0.85

d) 0.063 0.063


v) Checking Against Max. Shear Force on Back Wall in Horizontal Strip at Bottom Level.

a) The Maximum Shear Force occurs at Botton Level of Back Wall on its 53.475 kN/mHorizontal Strip which is also the Ultimate Shearing Force for the Section. Thus 53.475*10^3 N/m

b)

b-i) 1,000.000 mm


208.800 mm 216.000 mm

= Af-12.b/spro



Mpro


Relation between Provided Resisting Moment Mpro and Designed Moment MU.








VU.

Maximum Shear Force, VMax = VS-H-BW.= VU






h = tBack-Wall Depth of Well Cap.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h

b-iii) f 0.90

b-iv) - N

c) 1,134.000 kN/m 1134.000*10^3 N/m

1,643.128 kN/m 1643.128*10^3 N/m

1,134.000 kN/m 1134.000*10^3 N/m


c-ii) - N/m

c-iii) b 2.000

c-iv) - N

d) Vn>Vu Satisfied


f)Bottom Section does not Require any Shear Reinforcement, thus Flexural Design of Horizontal Reinforcement on


a) 62.138 N-mmwhere; 62.138*10^6 kN-m

69.042 N-mm 69.042*10^6 kN-m

f 0.90




The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-BW-H of Equations as mentioned in Aritical 5.8.3.3 :














Earth Face ofBack Wall is OK.




b)


d) Since Back Wall in Horizontal Direction is being considered as Fixed End 69.042 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 69.042*10^6 N-mm

e) 21.329 kN-m

21.329*10^6 N-mm

f) Mr>Ms-h-bw-usd Satisfied


a)

Where;

b) 75.343

13.179 kN-m 13.179*10^6 N-mm

753.982


c) 234.726


A 16,800.000 by Dividing the Total Concrete Area bounded in between Extreme Tension Face & a Straight Line parallel to Neutral Axis of Component having equal distance from



y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-H-BW

at its Fixed Ends will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment MU at Fixed Ends of assumed Rectangular (-)åMS-H-BW-USD

Beam in Horizontal on Earth Face = (-)åMS-H-BW-USD


Factored Moment M at Bottom of Vertical Span Strip ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied).



fs-Dev. is Developed Tensile Stress in Provided Reinforcements of Section under fs-Dev. N/mm2


i) M is Calculated Moment for the Section under Service Limit State (-)åMS-H-BW-WSD








Abutment Wall, CCov-Earth = 75mm & Bar Dia, DBar = 16f ; thus dc = (16/2 + 50)mm


the Centrioed of Main Tension Reinforcement Bars on both side & Diving the Area by the total Number of Main Bars as Tensile Reinforcement having Max. Clear


23,000.000 N/mm


246.000

d)

e) 7,382.546 N/mm




i)

Width Parameter, thus Provisions of Tensile Reinforcement in Horizontal on Back Wall Earth Surface in respect

j)

16Horizontal Span Strip :


a) The Calculated Flexural (+) ve Moment in Horizontal Span Strip of Back 8.532 kN-m/m

8.532*10^6 N-mm/m

Moment for Provision of Reinforcement against (+)ve Moment value. For (+)ve 51.966 kN-m/mvalue the required Reinforcement will be on Open Face of Back Wall. 51.966*10^6 N-mm/m

b) 51.966 kN-m/m

Cover = 50mm.In Backt Wall the Tension Bars in One Layer & as per Condition








Abutment Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.





Since Developed Tensile Stress of Tension Reinforcement of Back Wall, fs-Dev.< fs Allowable Tensile Stress;

the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allowable Max. Crack

of Control of Cracking & Distribution of Reinforcement are OK.


Flexural Design of Horizontal Reinforcements on Open Face of Back Wall against the (+) ve Moment on

(+)åMS-H-BW-USD


Mr

Since (+)åMS-H-BW-USD< Mr, the Allowable Minimum Moment for the Section, MU

51.966*10^6 N-mm/m


a) 12 mmWall.

b) 113.097

c) The provided Effective Depth for the Section with Reinforcement on Open 244.000 mm

d) 12.238 mm

e) 592.021

f) 191.036 mm,C/C

g) 150 mm,C/C

h) 753.982

i) 0.003

j) 17.318 mm

k) Resisting Moment for the Section with provided Steel Area, 72.752 kN-m/m

l) Mpro>Mu OK

m) pmax>ppro OK


a) 0.450

b) c 14.721 mm

c) 0.85


Let provide 12f Bars as Horizontal Reinforcement on Open Face of Back DBW-Open-H.

X-Sectional of 12f Bars = p*DAb-Open-V2/4 Af-12. mm2

de-pro.

Face, dpro = (tBack-Wall.-CCov-Open. -DBW-Open-V - DBW-Open-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-BW-Open-H. mm2/m

Spacing of Reinforcement with 12f bars = Af-12b/As-req-BW-Open-H. sreq


spro = 150mm,C/C

The provided Steel Area with 12f bars having Spacing 150mm,C/C As-pro-BW-Open-H. mm2/m

= Af-12.b/spro



Mpro


Status of Provided Resisting Moment Mpro in respect of Designed Moment MU.

Status of Provided Steel Ration rpro in respect of Allowable Max. Steel Ratio rMax.





d) 0.060


iv) Since Checking have been done for Provision of Horizontal Reinforcement on Back Wall Earth Face against Max. Moment due to Imposed Loads & found Satisfactory in all respect thus it does not require further Checking against Provision of Horizontal Reinforcement on Open Face of Back Wall.



Maximum

³ Mr)

mm,C/C

mm,C/C

Satisfied

Not Satisfy

mm,C/C

mm,C/C

Satisfied

mm,C/C

mm,C/C

K. Structural Design of Abutment Cap (Girder Seat) for Bridge:

1 Sketch Diagram of Abutment Cap.

2 Structural Data :

Description Notation. Value. Unit.


















SL

SAddl.

LGir.

WCarr-Way.

WS-Walk.

WCurb.

WR-Post.

WB-Deck.

RW&D.

PW&B.

hR-Post.

hCurb.

Rnos.

C/CD-R-Post.

tSlab.

tWC

35 035 0

30 0

700

60 0

300

2147

3047

450

150400

250 200

L1 = 9.350m

45 0 45 0

2000 2000 2000 2000 675675

10250

1200

2147

60 0

2147








ii) Dimentions of Abutment Cap & Back Wall :

a) Height of Back Wall 2.147 m

b) Thickness of Back Wall 0.300 m

c) Thickness of Wing Wall 0.450 m

d) Length of Back Wall between the Wing Walls 9.350 m

e) Length of Abutment Cap between outer Faces of Wing Walls 10.250 m

f) Height of Abutment Cap Rectangular Portion (Abutment Seat) 0.600 m

g) Height of Abutment Cap Trapezoid Portion (Abutment Seat) 0.300 m

h) Width of Abutment Cap Rectangular Portion (Abutment Seat) 1.000 m

i) Width of Abutment Cap Trapezoid Portion at Bottom(Abutment Seat) 0.450 m

j) Offset of Trapezoid Portion on Earth Face 0.400 m

k) Offset of Trapezoid Portion on Water Face 0.150 m

l) Span Length between Two Adjacent Bearings 2.000 m

m) Span Length between Exterior Bearing Center to Edge of Abutment Wall 1.125 m

2 Design Data :




vi)

9.807






vii)



NGirder.

NX-Girder.

hGirder.

hX-Girder.

bGirder.

bX-Girder.

C/CD-Girder.

hBack-Wall

tBack-Wall

tWing-Wall

LBack-Wall

LAb-Cap

hAb-Cap-Rec

hAb-Cap-Tro.

bAb-Cap-Rec

bAb-Cap-Tro-Bot.

bOffset-Earth.

bOffset-Water.

LInt-Ab-Cap.

LExt-Ab-Cap



gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3





a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa

ix) Design Data for Resistance Factors for Conventional Construction (AASHTO LRFD-5.5.4.2.1). :










j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85

x) Other Design Related Data :






wWater-Nor. kN/m3

wWater-Sali. kN/m3

wEatrh kN/m3


c



c Ec

= 0.043*24^(1.50)*21^(1/2) Mpa, (AASHTO LRFD-5.4.2.4).









fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.



VWL-Nor.

VWL-Spe.

VWA

a)

Here:


1.000

1.000

1.000



of 90 km/hr) for normal wind load only are selected as CRITICAL conditions for bridge structure.









ii) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1;










Qi = Force Effect,



but those would be occasional. Thus the respective Multiplier Factors of Limit State STRENGTH I (Bridge used by Normal Vehicle without wind load) for normal operation, Limit State of STRENGTH-III (Wind Velocity exceeding90km/hr) for wind load during cyclonic storm condition and Limit State of STRENGTH-IV (Having Wind Velocity



gEH


gEV


gES


Table 3.4.1-1&2 :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000






q) -

r) -



















t) 1.000












b) 1.000

c) IM 1.000 AASHTO LRFD-3.6.2.1, Table 3.6.2.1-1; SERVICE - I(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000




gEH


gEV


gES









i) 1.000



k) 1.000






q) -

r) -

t) 1.000

6 Computation of Imposed Load (DL & LL) upon Abutment Cap :

i) Philosophy of Flexural Design & Computation of different Loads (DL & LL) upon Well Cap:

a) Since the Abutment Cap will have to receive all Vertical Loads both from Live Load Elements & Dead Loads from the Superstructure and its Self Weight, thus computation of these Loads should done accordingly.

b)

Cap & Back Wall. Based on the mentioned assumptions of Uniformly Distributed Laod Reactions, Abutment Cap will be a Inverted Continuous Rectangular Beam having Girders as Reaction/Supports. Thus eacg Girder will haveenhance Reaction Values according their Position in respect of Calculated Uniformely Distributed Load Reaction upon the Abutment Wall.

c) Accordingly the Abutment Cap will be Designed as a Continuous Beam Having Uniformly Distributed Load upon it.

ii) Unfactored Dead Load Reactions from 1 no. Exteriod Girder :













Loads from Superstructure (DL & LL) will act upon Abutment Cap through the Girder Bearings as Concentrated Reactions. The Abutment Cap will also have to carry Dead Loads (DL) due to its Self Weight & Back Wall. Thesetotal Loads (DL & LL)will ultimately transmit to the Abutment Wall & those will behave as an Uniformly Distributed Upward Reaction against the Imposed Downward Loads (DL & LL) from Superstructure, Self Weight of Abutment

ii) Unfactored Dead Load Reactions from 1 no. Exteriod Girder :

a) Super-structural Dead Load without Wearing Course & Utilities 450.566 kN

b) Wearing Course & Utilities 40.031 kN

c) Total Dead Load for 1 no. Exterior Girder 490.598 kN

iii) Unfactored Dead Load Reactions from 1 no. Interiod Girder :

a) Super-structural Dead Load without Wearing Course & Utilities 358.382 kN


c) Total Dead Load for 1 no. Interior Girder 401.507 kN

iv) Unfactored Live Load Reactions from 1 no. Exteriod Girder :

a) Live Load due to HS20-44 Truck on Exterior Girder. 82.544 kN

b) 109.784 kN

c) 82.544 kN

d) Live Load due to Lane Load on Deck 25.187 kN

e) Live Load due to Pedistrian on Exterior Girder 56.250 kN

f) 191.221 kNunder Strength Limit State.

v) Unfactored Live Load Reactions from 1 no. Interiod Girder :

a) Live Load due to HS20-44 Truck on Interior Girder. 0.000 kN

b) 0.000 kN

c) 0.000 kN

d) Live Load due to Lane Load on Deck 77.500 kN

e) 77.500 kN

DLExt-Self & Sup-Imp.-UF

DLExt-WC&Uti-UF

DLExt-Total-UF

DLInt-Self & Sup-Imp.-UF

DLInt.-WC&Uti-UF

DLInt.-Total-UF

LLExt-Truck.-UF

Truck Live Load with Dynamic Load Allowance (IM) under Strength Limit LLExt-Truck+IM-UF

State = IM*LLExt-Truck.

Truck Live Load with Dynamic Load Allowance (IM) under Service Limit LLSExt-Truck.-IM-UF


LLExt-Lane.-UF

LLExt-Ped.-UF

Total Live Load for 1 no. Exterior Girder with Dynamic Load Allowance (IM) LLExt.-Total-UF

LLInt-Truck.-UF

Truck Live Load with Dynamic Load Allowance (IM) under Strength Limit LLInt-Truck+IM-UF

State = IM*LLInt-Truck.

Truck Live Load with Dynamic Load Allowance (IM) under Service Limit LLSInt-Truck.-IM-UF


LLInt-Lane.-UF

Total Leve Load for 1 no. Interior Girder with Dynamic Load Allowance (IM) LLInt.-Total-UF

under Strength Limit State.

vi) Factored Dead Load Reactions from 1 no. Exteriod Girder under Strength Limit State Design (USD):

a) Self Weight & Superimposed but without Wearing Course & 563.208 kN


c) 623.255 kN

vii) Factored Dead Load Reactions from 1 no. Interiod Girder under Strength Limit State Design (USD) :



c) 512.665 kN

viii) Factored Live Load Reactions from 1 no. Exteriod Girder under Strength Limit State Design (USD) :

a) 192.122 kN

b) Live Load due to Lane Load on Deck 44.078 kN

c) Live Load due to Pedistrian on Exterior Girder 98.438 kN

d) 334.638 kN

ix) Factored Live Load Reactions from 1 no. Interiod Girder under Strength Limit State Design (USD) :



c) 135.625 kN

x) Factored Dead Load Reactions from 1 no. Exteriod Girder under Service Limit State Design (WSD):

FDLExt-Self & Sup-Imp.-USD

Utilities = gDC*DLExt-Self & Sup-Imp.

FDLExt-WC&Uti-USD

= gDW*DLExt-WC&Uti

Total Factored Factored (USD) Dead Load (DL) for 1 no. Exterior Girder FDLExt-Total-USD

FDLInt-Self & Sup-Imp.-USD

Utilities = gDC*DLInt-Self & Sup-Imp.

FDLInt.-WC&Uti-USD

= gDW*DLInt-WC&Uti

Total Factored Dead Load (DL) for 1 no. Interior Girder FDLInt.-Total-USD

Live Load due to HS20-44 Truck on Exterior Girder + IM. FLLExt-Truck+IM.-USD

= m*IM*gLL-Truck*LLExt-Truck+IM.

FLLExt-Lane.-USD

= mgLL-Lane*LLExt-Lane.

FLLExt-Ped.-USD

= mgLL-PL.*LLExt-Ped.

Total Factored (USD) Leve Load (LL) for 1 no. Exterior Girder FLLExt.-Total-USD

FLLInt-Truck.+IM-USD

= m*IM*gLL-Truck*LLInt-Truck+IM.

FLLInt-Lane.-USD

= mgLL-Lane*LLInt-Lane.

Total Factored ((USD) Leve Load (LL)for 1 no. Interior Girder FLLInt.-Total-USD



c) 490.598 kN

xi) Factored Dead Load Reactions from 1 no. Interiod Girder under Service Limit State Design (WSD) :



c) 401.507 kN

xii) Factored Live Load Reactions from 1 no. Exteriod Girder under Service Limit State Design (WSD) :

a) 82.544 kN


c) Live Load due to Pedistrian on Exterior Girder 56.250 kN

d) 163.982 kN

xiii) Factored Live Load Reactions from 1 no. Interiod Girder under Service Limit State Design (WSD) :



c) 77.500 kN

xix) Loads due to Self Weight of Back Wall & Abutment Cap on Abutment Wall for per Meter Length(Considering Weight of Back Wall & Abutment Cap is Acting as Uniformly Distributed Loads.) :

a) 15.458 kN/m

FDLExt-Self & Sup-Imp.-WSD

Utilities = gDC*DLExt-Self & Sup-Imp.

FDLExt-WC&Uti-WSD

= gDW*DLExt-WC&Uti

Total Factored (WSD) Dead Load (DL) for 1 no. Exterior Girder FDLExt-Total-WSD

FDLInt-Self & Sup-Imp.-WSD

Utilities = gDC*DLInt-Self & Sup-Imp.

FDLInt.-WC&Uti-WSD

= gDW*DLInt-WC&Uti

Total Factored (WSD) Dead Load (DL) for 1 no. Interior Girder FDLInt.-Total-WSD

Live Load due to HS20-44 Truck on Exterior Girder + IM. FLLExt-Truck+IM.-WSD

= m*IM*gLL-Truck*LLExt-Truck

FLLExt-Lane.-WSD

= mgLL-Lane*LLExt-Lane.

FLLExt-Ped.-WSD

= mgLL-PL.*LLExt-Ped.

Total Factored (WSD) Leve Load (LL) for 1 no. Exterior Girder FLLExt.-Total-WSD

FLLSInt-Truck.+IM

= m*IM*gLL-Truck*LLInt-Truck

FLLSInt-Lane.

= mgLL-Lane*LLInt-Lane.

Total Factored (WSD) Leve Load (LL)for 1 no. Interior Girder FLLInt.-Total-WSD

Dead Load due to Back Wall, = tBack-Wall*hBack-Wall*wc DLBack-Wall

b) 14.400 kN/m

c) Total Unfactored Uniformly Distributed Load per meter on Abutment 29.858 kN/m Wall due to Back Wall & Abutment Cap.

d) Total Factored Dead Load due to Back Wall & Abutment Cap under 37.323 kN/m

e) Total Factored Dead Load due to Back Wall & Abutment Cap under 29.858 kN/m

xx)State Design (USD) :

a) 957.892 kN

b) 648.290 kN

xxi)State Design (WSD) :

a) 654.579 kN

b) 565.489 kN

6 Sketch Diagram of Actual Imposed Factored Loads (DL & LL) from Girder. Abutment Cap & Back Wall Under Strength Limit State (USD) :

957.892 648.29 648.2898438 648.29 957.892

kN kN kN kN kN

2.147 m

1.125 2.00 2 2 2.00 1.125

m m m m m m

Self Wt. Of Aburment Cap + Back Wall 0.600 m

37.323 kN/m

0.300 m

10.250 m

Dead Load due to Self Weight of Abutment Cap Rectangular Portion DLAb.-Cap-Rec.

= bAb-Cap-Rec*hAb-Cap-Rec.*wc

DLBack+Ab.Cap.

FDLBack+Ab.Cap.-USD

Strength Limit State Design (USD) = gDC*(DLBack+Ab.-Cap)

FDLBack+Ab.Cap.WSD

Service Limit State Design (WSD) = gDC*(DLBack+Ab.-Cap)

Total Factored Load Reactions (DL + LL) from 1 no. Exterior & 1 no. Interiod Girder under Strength Limit

Total Factored Load Reactions (DL + LL) from 1 no. Exterior Girder FLExt.-Total(DL+LL)-USD

= FDLExt-Total-USD + FLLExt.-Total-USD

Total Factored Load Reactions (DL + LL) from 1 no. Interior Girder FLInt.-Total(DL+LL)-USD

= FDLInt-Total-USD + FLLInt.-Total-USD

Total Factored Load Reactions (DL + LL) from 1 no. Exterior & 1 no. Interiod Girder under Service Limit

Total Factored Load Reactions (DL + LL) from 1 no. Exterior Girder FLExt.-Total(DL+LL)-WSD

= FDLExt-Total-WSD + FLLExt.-Total-WSD

Total Factored Load Reactions (DL + LL) from 1 no. Interior Girder FLInt.-Total(DL+LL)-WSD

= FDLInt-Total-WSD + FLLInt.-Total-WSD

RGir.-Ext RGir.-Int RGir.-Int RGir.-Int RGir.-Ext.

hBack-Wall =

hAb-Cap-Recl =

FDLBack+Ab.Cap.=

hAb-Cap-Tro =

LAb-Cap =

45 0 45 0

30 0

700

7 Computation of Unfactored Uniformly Distributed Dead Loads for Abutment Cap and Resultant Reactionson Girder Points :

i) Computation of Uniformity Distributed Dead Loads upon Abutment Cap :

a) 981.195 kN

b) 1204.521 kN

c) 2185.716 kN

d) 213.241 kN/m its per Meter Length due to Imposed Loads from Superstructure through Girders

e) 29.858 kN/m

f) 243.099 kN/mCap for its per Meter Length due to Imposed Loads from Superstructure through

ii) Resultant Reaction on Girders against Computed of Uniformity Distributed Dead Loads :

a) 516.585 kN

b) 486.198 kN

8 Computation of Uniformty Distributed Factored Loads (DL & LL) on Abutment Cap & Resultant Reactionsfor the Girder Points Under Strength Limit State (USD) :

Unfactored Imposed Dead Loads (DL) from 2nos. Exterior Girders upon UFDL-Ext.-Gir.

Abutment Cap = 2*DLExt.-Total-UF

Unfactored Imposed Dead Loads (DL) from 3nos. Interior Girders upon UFDL-Int.-Gir.

Abutment Cap = 3*DLInt.-Total-UF

Total Unfactored Dead Loads (DL) upon Abutment Cap from Superstructure UFDL-Super.

= (UFLExt.-Gir, + UFLInt.-Gir.)

Uniformly Distributed Unfactored Dead Loads (DL) upon Abutment Cap for wDL-Super.-UF

= UFDL-Super,/ LAb-Cap.

Uniformly Distributed Unfactored Dead Loads (DL) upon Abutment Cap wDL-Back+Ab.Cap.-UF

for its Self Weight & Back Wall = DLBack+Ab.Cap.

Total Uniformly Distributed Unfactored Dead Loads (DL) upon Abutment wDL-Ab-Cap.-UF

Girders, its Self Weight & Back Wall = (wDL-Super,-UF + wDL-Back+Ab.Cap-UF.)

Reaction on Each Exterior Girder = wDL-Ab-Cap.*(LExt-Ab-Cap + LInt-Ab-Cap./2) RDL-Ext

Reaction on Each Interior Girder = wDL-Ab-Cap.*LInt-Ab-Cap. RDL-Int

35 035 0

700

60 0

300

2747

3047

15 0

400

2147

450

A B

D C

Load fromSuperstructure

i) Computation of Uniformty Distributed Factored Loads (DL & LL) upon Abutment Cap :

a) 1915.784 kN

b) 1944.870 kN

c) 3860.654 kN

d) 376.649 kN/mper Meter Length due to Imposed Loads from Superstructure through Girders

e) 37.323 kN/m

f) 413.972 kN/mfor its per Meter Length due to Imposed Loads from Superstructure through

ii) Resultant Reaction on Girders against Computed of Uniformity Distributed Factored Loads (DL & LL):

a) 879.691 kN

b) 827.944 kN

9 Computation of Uniformty Distributed Unfactored Loads (DL & LL) on Abutment Cap & Resultant Reactionsfor the Girder Points Under Service Limit State (WSD) :

i) Computation of Uniformty Distributed Factored Loads (DL & LL) upon Abutment Cap :

a) 1309.159 kN

b) 1437.021 kN

c) 2746.179 kN

d) 267.920 kN/mits per Meter Length due to Imposed Loads from Superstructure through Girders

Factored Imosed Loads (DL & LL)from 2nos. Exterior Girders upon FLExt.-Gir.-USD

Abutment Cap = 2*(FDLExt.-Total-USD + FLLExt.-Total-USD)

Factored Imosed Loads (DL & LL)from 3nos. Interior Girders upon FLInt.-Gir.-USD

Abutment Cap = 3*(FDLInt.-Total-USD + FLLInt.-Total-USD)

Total Factored Loads (DL & LL) upon Abutment Cap from Superstructure FLSuper.-USD

= (FLExt.-Gir,-USD + FLInt.-Gir-USD.)

Uniformty Distributed Factored Loads (DL & LL) upon Abutment Cap for its wSuper.-USD

= FLSuper,-USD/ LAb-Cap.

Uniformty Distributed Factored Loads (DL & LL) upon Abutment Cap wBack+Ab.Cap.-USD

for its Self Weight & Back Wall = FDLBack+Ab.Cap-USD

Total Uniformty Distributed Factored Loads (DL & LL) upon Abutment Cap wAb-Cap.-USD

Girders, its Self Weight & Back Wall = (wSuper,-USD + wBack+Ab.Cap.-USD)

Reaction on Each Exterior Girder = wAb-Cap.-USD*(LExt-Ab-Cap. + LInt-Ab-Cap./2) RExt-USD

Reaction on Each Interior Girder = wAb-Cap.-USD*LInt-Ab-Cap. RInt-USD

Unfactored Imosed Loads (DL & LL) from 2nos. Exterior Girders upon FLExt.-Gir.-WSD

Abutment Cap = 2*(FDLExt.-Total-WSD + FLLExt.-Total-WSD)

Unfactored Imosed Loads (DL & LL)from 3nos. Interior Girders upon FLInt.-Gir.-WSD

Abutment Cap = 3*(FDLInt.-Total-WSD + FLLInt.-Total-WSD)

Total Unfactored Loads (DL & LL) upon Abutment Cap from Superstructure FLSuper.-WSD

= (FLExt.-Gir,-WSD + FLInt.-Gir.-WSD)

Uniformty Distributed Unfactored Loads (DL & LL) upon Abutment Cap for wSuper.-WSD

= FLSuper,-WSD/ LAb-Cap.

e) 29.858 kN/m

f) 297.778 kN/mfor its per Meter Length due to Imposed Loads from Superstructure through

ii) Resultant Reaction on Girders against Computed of Uniformity Distributed Factored Loads (DL & LL):

a) 632.779 kN

b) 595.557 kN

10 Sketch Diagram of Computed Factored Loads (DL & LL) on Abutment Cap & Resultant Reactions :

879.691 827.94 827.9443289 827.94 879.691

kN kN kN kN kN

2.147 m

1.125 2.000 2.000 2.000 2.000 1.125

m m m m m m

Computed Uniformly Distributed Loads on Aburment Cap 0.600 m

413.972 kN/m

0.300 m

10.250 m

11 Calculations for Moments under different obstions of Computed Loads from Superstructure, Self Weight of Abutment Cap & Back Wall on Abutment Cap Section 'ABCD' :

i) Calculations for Factored Moments at different Reaction Positions Under Strength Limit State (USD):

Uniformty Distributed Unfactored Loads (DL & LL) upon Abutment Cap wBack+Ab.Cap.-WSD

for its Self Weight & Back Wall = FDLBack+Ab.Cap.-WSD

Total Uniformty Distributed Unfactored Loads (DL & LL) upon Abutment Cap wAb-Cap.-WSD

Girders, its Self Weight & Back Wall = (wSuper-WSD, + wBack+Ab.Cap-WSD.)

Reaction on Each Exterior Girder = wAb-Cap.-WSD*(LExt-Ab-Cap. + LInt-Ab-Cap./2) RExt-WSD

Reaction on Each Interior Girder = wAb-Cap.-WSD*LInt-Ab-Cap. RInt-WSD

RExt-USD. RInt-USD.. RInt-USD. RInt.-USD. RExt-USD.

hBack-Wall =

hAb-Cap-Recl =

wF-Ab-Cap.-USD =hAb-Cap-Tro =

LAb-Cap =

35 035 0

30 0

700

60 0

300

2747

3047

15 0

400

2147

450

A B

D C

Load fromSuperstructure

45 0 45 0

a) (-) ve Moments on Outer Face of Exterior Girder Reaction Position, 261.967 kN-m

b) (-) ve Moments at Inner Face of Exterior Girder Reaction Position, 165.589 kN-m

c) (-) ve Moments at Face of Interior Girder Reaction Position, 183.988 kN-m

d) (+) ve Moments at Middle Position in-between Two Reactions, 206.986 kN-m

ii) Calculations for Factored Moments at different Reaction Positions Under Service Limit State (WSD):





iii) Calculations for Dead Load (DL) Moments at different Reaction Positions:





11 Calculations for Shearing Forces under different options of Computed Loads from Superstructure, Self Weight of Abutment Cap & Back Wall on Abutment Cap Section 'ABCD' :

i) Factored Shearing Forces at different Reaction Positions Under Strength Limit State (USD):

(-) MExtr-Out-USD

= wAb-Cap.-USD*LExt-Ab-Cap.2/2

(-) MExtr-Inner-USD

= wAb-Cap-USD.*LInt-Ab-Cap.2/10

(-) MInte-Face-USD

= wAb-Cap.USD*LInt-Ab-Cap.2/9

(+) MMid-USD

= wAb-Cap.-USD*LInt-Ab-Cap.2/8

(-) MExt-Out-WSD

= wAb-Cap.-WSD*LExt-Ab-Cap.2/2

(-) MExt-Int-WSD

= wAb-Cap.-WSD*LInt-Ab-Cap.2/10

(-) MInt-WSD

= wAb-Cap.-WSD*LBearing2/9

(+) MMid-WSD

= wAb-Cap.-WSD*LInt-Ab-Cap.2/8

(-) MDL-Extr-Out

= wDL-Ab-Cap.*LExt-Ab-Cap.2/2

(-) MDL-Extr-Inner

= wDL-Ab-Cap.*LInt-Ab-Cap.2/10

(-) MDL-Interior


(+) MDL-Middle


a) Shear Forces on Exterior Girder Reaction Position Outer Face, 465.719 kN

b) Shear Forces on Exterior Girder Reaction Position Inner Face, 413.972 kN

c) Shear Forces on Interior Girder Reaction Position of Both Faces, 413.972 kN

d) Shear Forces on Mid between Interior Girder Reaction Positions 0.000 kN

ii) Factored Shearing Forces at different Reaction Positions Under Service Limit State (WSD):

a) Shear Forces at Exterior Girder Reaction Position Outer Face, 335.001 kN

b) Shear Forces at Exterior Girder Reaction Position Inner Face, 297.778 kN

c) Shear Forces at Interior Girder Reaction Position of Both Faces, 297.778 kN


iii) Calculations for Dead Load Shearing Forces at different Reaction Positions:

a) Shear Forces at Exterior Girder Reaction Position Outer Face, 273.486 kN

b) Shear Forces at Exterior Girder Reaction Position Inner Face, 273.486 kN

c) Shear Forces at Interior Girder Reaction Position of Both Faces, 243.099 kN


11 Computation of Related Features required for Flexural Design of Main Reinforcements for Abutment Cap Against the Calculated Bending Moments on its Different Sections :

i) Provisions for Reinforcements, Design Width-Depth & Clear Cover on different Faces of Abutment Cap :

a) Against (-) ve Moment values the Abutment Cap Requires the Reinforcements on its Top Surface, whereas for the

(-)VR-Extr-Out-USD.

= wAb-Cap.-USD.*LExt.-Ab-Cap.

VR-Extr-Inn-USD.

= RExt.-USD. - VR-Ext.-Out-USD.

VR-Int-Int-USD.

= RInt.-USD. - wAb-Cap.-USD.*LInt-Ab-Cap./2

VMid-of-Int-USD.

= VR-Int-Int.-USD. - wAb-Cap.-USD.*LInt-Ab-Cap./2

(-)VR-Extr-Out-WSD

= wAb-Cap.-WSD*LExt.-Ab-Cap.

VR-Extr-Inn.-WSD

= RExt.-WSD - VR-Extr-Out-WSD

VR-Int-Int.-WSD

= RInt.-WSD - wAb-Cap.-WSD*LInt-Ab-Cap./2

VMid-of-Int.-WSD

= VR-Int-Int-WSD - wAb-Cap.-WSD*LInt-Ab-Cap./2

(-)DLVR-Extr-Out

= wDL-Ab-Cap.*LExt.-Ab-Cap.

DLVR-Extr-Inn.

= RDL.Ext. - DLVR-Extr-Out.

DLVR-Int-Int.

= RDL.Int. - wDL-Ab-Cap.*LInt-Ab-Cap./2

DLVMid-of-Int.

= DLVR-Int-Int.. - wUF-Ab-Cap.*LInt-Ab-Cap./2

(+) ve Moment values the Reinforcements will on its Bottom Surface.

b) h 0.6 m

c) b 1 m

d) 75 mm

e) Let the Clear Cover on Top, Face of Abutment Cap = 50mm, 50 mm

f) Let the Clear Cover on Bottom Face of Abutment Cap = 50mm, 50 mm

g) Let the Clear Cover on Water Face of Abutment Cap = 50mm, 50 mm



b) c Variable

c) Variable

Variable

Variable

410.00

Variable

Variable mm

Variable mm






Height on Rectangular Portion of Abutment Cap is Design Depth = hAb-Cap-Rec

Height on Rectangular Portion of Abutment Cap is Design Width = bAb-Cap-Rec

Let the Clear Cover on Earth Face of Abutment Cap = 75mm, C-Cov-Earth.

C-Cov-Top

C-Cov-Bot.

C-Cov-Water

c/de-Max.











Centroid in mm.




thus de = ds .






Variable N-mm

Variable

0.060000

60.000/10^3

60.000*10^6

2.887

c) 173221361.269 N-mm

173.221 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)




Abutment 5.7.3.3.2-1 M (1.33*M)Cap kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

153.836 173.221 173.221 173.221 207.866 261.967 348.416 207.866 261.967

(-)ve on 97.240 173.221 173.221 173.221 207.866 165.589 220.233 207.866 207.866

(-)ve on 108.044 173.221 173.221 173.221 207.866 183.988 244.704 207.866 207.866

121.549 173.221 173.221 173.221 207.866 206.986 275.291 207.866 207.866

iv)










Snc = (b*h.3/12)/(h./2)










for RCC Mu


(-)ve on

RExt-Outer

RExt-Inner

RInt-Inner

(+)ve on RInt-Mid


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),

b) 0.016

12 Flexural Design of Reinforcements on Top Surface of Abutment Cap against the Max. (-) ve Moment :


a) The Maximum Flexural (-) ve Moment occurs at Exterior Position of 261.967 kN-m/mAbutment Cap. Against Calculated Factored (-) ve Moment value 261.967*10^6 N-mm/m

the Provision of Reinforcements will be on Top Surface of Abutment Cap. 207.866 kN-m/m 207.866*10^6 N-mm/m

b) 261.967 kN-m/m

261.967*10^6 N-mm/m

ii) Provision of Reinforcement on Top Surface of Abutment Cap :

a) 25 mmCap.

b) 490.874

c) The provided Effective Depth for the Section with Reinforcement on Top 537.500 mm

d) 28.035 mm

e) 1,356.181

f) 2.763 nos.

g) 8 nos.Abutment Cap.

h) 3,926.991


a) 0.007

b) 90.200 mm



(-) MF-Extr-Out-USD

Mr

Since (-) MF-Extr-Out-USD > Mr, the Allowable Minimum Moment for the Section, MU

thus (-) MF-Extr-Out-USD is the Design Moment MU.

Let provide 25f Bars as Main Reinforcement on Top Surface of Abutment DAb-Cap-Top

X-Sectional of 25f Bars = p*DAb-Cap-Top2/4 Af-25. mm2

de-pro-Top.

Surface, dpro = (h -CCov-Top -DAb-Cap-Top/2)

With Design Moment MU , Design Width of Cap b & Effective Depth dpro; areq.


2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Ab-Cap-Top mm2/m

Number of 25f bars Requred for the Section= As-req-Ab-Earth-V./Af-20 NBer-req.

Let the Provided 8 nos. 25f bars as Main Reinforcement on Top Surface of NBer-pro.

The provided Steel Area with 8 nos. 25f bars as Main Reinforcement As-pro-Ab-Cap-Top mm2/m

on Top Surface of = Af-25.*NBer-pro.



Mpro


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 76.670 mm

c) 0.85

d) 0.143


v) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 713.517 N-mmwhere; 713.517*10^6 kN-m

792.797 N-mm 792.797*10^6 kN-m

f 0.90

b)


d) Since Abutment Cap is being considered as a Continous Rectangular Beam 792.797 kN-mhaving Cantilver arm on both End. The Steel Area against Factored Max. (-) ve 792.797*10^6 N-mm

e) 261.967 kN-m

261.967*10^6 N-mm

f) Mr>Mu Satisfied















y(d/s-a/2)

Mn = Asfy(ds-a/2); here ds = dpro & a = apro.

Mn-Top

Moment is at its Top Surface for which the value of Nominal Resistance,

Mn = Asfy(ds-a/2).

Calculated Factored (-) ve Moments MU at Exterior Girder Posotions are (-)MF-Extr-Out-USD

with Maximum value = (-)MF-Extr-Out-USD


Factored Moment MU on Exterior Girder Position ( Which one is Greater, if Mr ³ MU the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

a)

Where;

b) 89.275

188.438 kN-m 188.438*10^6 N-mm

3,926.991


c) 231.825

62.500 mmTension Bar. The Depth is Summation Top Clear Cover & Radius of the


23,000.000 N/mm


246.000

d)

e) 8,857.200 N/mm





i) M is Calculated Moment for the Section under Service Limit (-)MF-Extr-Out-WSD





i) dc= Depth of Concrete Extreme Tension Face from the Center of Closest dc


of Rectangular Section, dc = DBar/2 + CCov-Top. Since Clear Cover on Top Face of

Abutment Cap, CCov-Top = 50mm & Bar Dia, DBar = 25f ; thus dc = (25/2 + 50)mm


Cover = 50mm. In Abutment Cap the Tension Bars in One Layer & as per Condition


Surrounding a Single Tension Bar can Compute by A = 2dc*b/NBar-pro.






Abutment Wall Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.





i)

Width Parameter, thus Provisions of Tensile Reinforcement on Abutment Cap Top Surface in respect of Control

j)

13


a) The Maximum Flexural (+) ve Moment occurs at Middle Position of 206.986 kN-m/mAbutment Cap Sections in-between Two Girders. Against Calculated Factored 206.986*10^6 N-mm/m

(+) ve Moment values the Reinforcements will be on Bottom Surface of 207.866 kN-m/mAbutment Cap. 207.866*10^6 N-mm/m

b) 207.866 kN-m/m

206.986*10^6 N-mm/m

ii) Provision of Reinforcement on Bottom Surface of Abutment Cap :

a) 25 mmAbutment Cap.

b) 490.874

c) The provided Effective Depth for the Section with Reinforcement on Bottom 537.500 mm

d) 22.121 mm

e) 1,065.531

f) 2.171 nos.

g) 8 nos.Abutment Cap.




Since Developed Tensile Stress on Abutment Cap Tension Reinforcement fs-Dev.< fsa Computed Tensile Stress;

the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev.< ZMax. allowable Max.Crack

of Cracking & Distribution of Reinforcement are OK.


Flexural Design of Reinforcements on Bottom Surface of Abutment Cap against (+) ve Moment :

(+) MInt-Mid-USD

Mr

Since (+) MInt-Mid-USD > Mr, the Allowable Minimum Moment for the Section, MU

thus (+) MInt-Mid-USD is the Design Moment MU.

Let provide 25f Bars as Main Reinforcement on Bottom Surface of DAb-Cap-Bot.

X-Sectional of 25f Bars = p*DAb-Cap-Bot2/4 Af-25. mm2

de-pro-Bot.

Surface, dpro = (h -CCov-Bot -DAb-Cap-Bot/2)

With Design Moment MU , Design Width of Cap b & Effective Depth dpro; areq.


2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Ab-Cap-Bot mm2/m

Number of 25f bars Requred for the Section= As-req-Ab-Earth-V./Af-20 NBer-req.

Let the Provided 6 nos. 25f bars as Main Reinforcement on Top Surface of NBer-pro.

h) 3,926.991


a) 0.007

b) 90.200 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 76.670 mm

c) 0.85

d) 0.143


v) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 713.517 N-mmwhere; 713.517*10^6 kN-m

792.797 N-mm 792.797*10^6 kN-m

f 0.90

b)


The provided Steel Area with 6 nos. 25f bars as Main Reinforcement As-pro-Ab-Cap-Bot mm2/m

on Top Surface of = Af-25.*NBer-pro.



Mpro


Relation between Provided Resisting Moment Mpro amd Calculated Design Moment MU.













y(d/s-a/2)

Mn = Asfy(ds-a/2); here ds = dpro & a = apro.

d) Since Abutment Cap is being considered as a Continous Rectangular Beam 792.797 kN-mhaving Cantilver arm on both End. The Steel Area against Factored Max. (-) ve 792.797*10^6 N-mm

e) 206.986 kN-m

206.986*10^6 N-mm

f) Mr>Mu Satisfied


a)

Where;

b) 70.538

148.889 kN-m148.889*10^6 N-mm

3926.990817


c) 231.825

62.500 mmTension Bar. The Depth is Summation of Bottom Clear Cover & Radius of the


Mn-Top

Moment is at its Top Surface for which the value of Nominal Resistance,

Mn = Asfy(ds-a/2).

Calculated Factored (+) ve Moments MU at Middle Posotion in-between Two MInt.-Mid.-USD

Girdres have the Maximum value = MInt.-Mid.-USD


Factored Moment MU on Exterior Girder Position ( Which one is Greater, if Mr ³ MU the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State (WSD) MInt.-Mid.-WSD





i) dc= Depth of Concrete Extreme Tension Face from the Center of Closest dc

Closest Bar to Tension Face with Max. Clear Cover = 50mm. In a Component

of Rectangular Section, dc = DBar/2 + CCov-Bot. Since Clear Cover on Bottom Face

of Abutment Cap, CCov-Bot = 50mm & Bar Dia, DBar = 25f ; thus dc = (25/2 + 50)mm


Cover = 50mm. In Abutment Cap the Tension Bars in One Layer & as per Condition


Surrounding a Single Tension Bar can Compute by A = 2dc*b/NBar-pro.

23,000.000 N/mm


246.000

d)

e) 6,998.282 N/mm

f) fs-Dev.< fsa Satisfy



i)

Width Parameter, thus Provisions of Tensile Reinforcement on Abutment Cap Top Surface in respect of Control

j)

14 Design of Shear Reinforcement for Abutment Cap against Shearing Forces & Checkings:

i)

a)

Table-1. For Shearing Forces at different Sections of Abutment Cap due to Factored Loads (DL & LL)

Searing On Outer On Inner On Inner Middl

Force Face of Face of Face of between

Locations Ext. Girder Ext. Girder Int. Girder Int. GirdersUnit kN kN kN kN

Shearing 465.719 413.9722 413.972 0.000

Force

b) Factored Shearing Forces on Outer Face of Exterior Girder. 465.719 kN

c) Factored Shearing Forces on Inner Face of Exterior Girder. 413.972 kN











Since Developed Tensile Stress on Abutment Cap Tension Reinforcement fs-Dev.< fs Allowable Tensile Stress;

the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev.< ZMax. allowable Max.Crack

of Cracking & Distribution of Reinforcement are OK.



Table for Max. Shear Forces at Different Locations of Abutment Cap due to Applied Factored Loads (DL & LL);

Vu-Ext.-Outer

Vu-Ext.-Inner

d) Factored Shearing Forces on Inner Face of Interior Girder. 413.972 kN

d) Factored Shearing Forces on Mid of Interior Girders. - kN


a)

b) - Mpa

c) 1.00 mm

d) Variable mm


0.72h 0.432 mm

Variable mm

h 0.600 mm

e) f 0.90

f)

Locations Width Depth of Effective Calculated Calculated Calculated Calculated

of of Abutment Depth of value of value of value of value of

Shearing Abutmrnt Cap Section for 0.9de 0.72h Effective Shearing

Force Width Rectangular Tensial Shear Depth Stress

(bv) (h)mm mm mm mm mm mm

On Outer 1000.000 600.00 537.50 483.75 432.00 483.75 1.070

Face of

Ext.-Girder

On Inner 1000.000 600.00 537.50 483.75 432.00 483.75 0.951

Face of

Ext.-Girder

On Inner 1000.000 600.00 537.50 483.75 432.00 483.75 0.951

Face of

Int.-Girder

On Middle 1000.000 600.00 537.50 483.75 432.00 483.75 0.000

between

Int.-Girders

Vu-Int-Inner.

Vu-Mid-of-Inner.




bv is Width of Abutment Cap = bAb-Cap-Rec.. bv


0.9de



ii) h is Depth of Abutment Cap Rectangular Section, hAb-Cap-Rec.


Table for Computation of values of vu, de, bv, 0.9de , 0.72h & dv at different Section of Abutment Cap.

Table-3 Values of vu, de, bv, 0.9de , 0.72h & dv at different Section of Abutment Cap.

(de) (dv) (vu)N/mm2

g) Shearing Stress due to Applied Factored Shearing Forces at Different Sections of Abutment Cap :

i) Shearing Stress due to Applied Shearing Force on Outer Face of Exterior 1.070

ii) Shearing Stress due to Applied Shearing Force on Inner Face of Exterior 0.951

iii) Shearing Stress due to Applied Shearing Force on Inner Face of Interior 0.951

iv) Shearing Stress due to Applied Shearing Force on Middle between -


a)

b) N

c) N

d) 0.90


a)

b)

c)

d)

e)

f)

g)

a 90

vExt.-Outer N/mm2

Girder = VExt.-Outer/fbvdv N/mm2

vExt.-Inner N/mm2

Girder = VExt.-Inner/fbvdv N/mm2

vInt.-Inner N/mm2

Girder = VInt.-Inner/fbvdv N/mm2

vMid-of-Inner. N/mm2

Interior Girders = VMid-of-Inner. /fbvdv N/mm2







The Nominal Shear Resistance Vn at any Section of Abutment Cap is the Lesser value Computed from Equations

i) Vn = Vc + Vs + Vp (Equ. 5.8.3.3-1) &









h)

i) - N

v)

a)



d)

e) -

f) -

g)

3,926.991

3,926.991

3,926.991

3,926.991

h) - MPa.multiplied by Locked-in differencein Strain between the Prestressing Tendonsand Surrounding Concrete (Mpa). For the usal level of Prestressing, the value


j)

k)




Computation of Value b & q at different Locations of Abutment Cap as per AASHTO-LRFD-5.8.3.4:

Calculation of Longitudinal Strain in Web Reinforcement εs in mm/mm on the Flexural Tension side of AbutmentCap according to Equations ;




Ac is Area of Concerte on Flexural Tention side of Abutment Cap in mm2 having valueAc mm2

Ac = bX-Web* hX-Gir,/2 mm2



As is Area of Non-Prestressing Steel on Flexural Tention side of Abutment Cap


i) On Outer Face of Exterior Girder the Provided Steel Area in mm2 As-Ext.-Outer. mm2

ii) On Inner Face of Exterior Girder the Provided Steel Area in mm2 As-Ext.-Inner. mm2

iii) On Inner Face of Interior Girder the Provided Steel Area in mm2 As-Int.-Inner. mm2

iv) On Middle of Interior Girders the Provided Steel Area in mm2 As-Mid.-of-Inner. mm2






Mu is Factored Moment quantity of the Section in N-mm but not less then Vudv.


I)

Locations Factored Factored Calculated Calculated Effective

of Moment Shear value of value of Moment

Shearing for the for the Effective value

Force Section. Section. Shear Depth

kN-m kN mm kN kN-m

On Outer 261.967 465.719 483.750 225.291 261.967

Face of

Ext.-Girder

On Inner 165.589 413.972 483.750 200.259 200.259

Face of

Ext.-Girder

On Inner 183.988 413.972 483.750 200.259 200.259

Face of

Int.-Girder

On Mid. 275.291 0.000 483.750 0.000 275.291

between

Int.-Girders

vi)

a) 0.051

b) 0.045

c) 0.045

d) -

vii)provisions of Equation 5.8.3.4.2-1 :

a)

b) 4.420

c)



Vudv

MuEff



Value of vc/f/c on Outer Face of Exterior Girder vExt.-Outer/f/

c

Value of vc/f/c on Inner Face of Exterior Girder vExt.-Inner/f/

c

Value of vc/f/c on Interior Girder Position vInt.-Inner/f/

c

Value of vc/f/c on Middle Position in-between Interior Girders vMid.of-Inner/f/

c


Since for RCC Compoent values of Prestressing Components Nu, Vp, Asp, fpo, Ep etc. are = 0, thus equation



Considering the Initial value of ex = 0.001on Exterior Girder Outer Face & with cotq

the Effective Moment Mu-Eff. From the Equation Cotq = (ex2EsAs - Mu/dv)/0.5Vu


d) 1.000

e) 0.800

f) 0.825

g) 0.362

j)

viii)

a)

Location of Abutment Cap Referece Value of

Section of Table q ba) On Exterior Girder Outer Face 5.8.3.4.2.-1 0.051 1.000 36.40 2.23

b) On Exterior Girder Outer Face 5.8.3.4.2.-2 0.045 0.800 36.40 2.23

c) On Interior Girder Inner Face 5.8.3.4.2.-2 0.045 0.825 36.40 2.23

b) On Middle of Interior Girders 5.8.3.4.2.-2 0.000 0.362 30.50 2.59

b)

1.356

1.356

1.356

1.698

ix)

a)


x) Regions Requiring Transverse or Shear/Web Reinforcements under AASHTO-LRFD-5.8.2.4 :


On Outer Face of Exterior Girder = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Ext-Outer*1000

On Inner Face of Exterior Girder = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Ext-Inner*1000

On Interior Girder Position = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Int-Inner*1000

On Mid of Interior Girders = ((Mu/dv + 0.5Vucotq )/(2EsAs))*1000 ex-Mid. -Of-Int*1000

Since at all Positions of the Abutment Cap posses values of es-x1000 > 0.50 ≤ 1.00 , thus values of q & b of these

Sections can obtain from Table -5.8.3.4.2-1 in respect of values of vc/f/c.

Computation of Values of q & b from AASHTO-LRFD'sTable -5.8.3.4.2-1. against the Respective Calculated

values of esx1000, Ratio vc/f/c :


vc/f/c exx1000


i) On Outer Face of Exterior Girder value of Cotq Cotq

ii) On Inner Face of Exterior Girder value of Cotq Cotq

iii) On Interior Girder Position value of Cotq Cotq

iv) On Mid of Interior Girders value of Cotq Cotq


Since Critical Section is at the Outer Face of Exterior Girder, thus Values of q & b of that Section are Governing

for Computation of Nominal Shearing Strength of Concrete (Vc) for the Abutment Cap.



b)


d) - N

e)

f) f 0.90

g)

h)

Location of Abutment Cap Values of Relation Equation

Section b Between Satisfied/

N N N Not Satisfied

a) On Exterior Girder Outer Face 2.230 465718.685 39,485.944 17768.675 Vu>0.5Vc Satisfied

b) On Exterior Girder Outer Face 2.230 413972.164 39,485.944 17768.675 Vu>0.5Vc Satisfied

c) On Interior Girder Inner Face 2.230 413972.164 39,485.944 17768.675 Vu>0.5Vc Satisfied

b) On Middle of Interior Girders 2.590 0.000 45,860.356 20637.160 Vu<0.5Vc Not Satisfied

i)

on Middle in-between Girders. Thus the Abutment Cap requires Transverse or Shear/Web Reinforcements.

xi)

a)

b) 405.670 kN 405.670*10^3 N

c) 405.670 kN 405.670*10^3 N

d) 405.670 kN 405.670*10^3 N

xii)

a)






f is Resistance Factor as per to AASHTO-LRFD-5.5.4.2. having value 0.90



Vu Vc 0.5fVc

Vu & Vc

The Table indicates that all the Sections of RCC Abutment Cap have satisfied the required provisions for Transverse/ Shear Reinforcements under Equation Vu > 0.5f (Vc + Vp). AASHTO-LRFD-Equation-5.8.2.4-1. Other than that




Vs = Avfydvcotq /s,

On Outer Faces of Exterior Girder, Vs= Avfydv-Ext-Gir-OuterCotq/spro. VS-Ext-Gir-Outer

On Inner Face of Exterior Girder, Vs=Avfydv-Ext-Gir-InnerCotq/spro. VS-Ext-Gir-Inner

On Position of Interior Girders, Vs=Avfydv-On-Inner-GirCotq/spro. VS-On-Inner-Gir

Computation of values for Nominal Shear Resistance (Vn) at Different Section of Abutment Cap under the

Provisions of AASHTO-LRFD-5.8.3.3 against Equation Vn = Vc + Vs + Vp (Equ. 5.8.3.3-2) :


b)

c) 445.156 kN 445.156*10^3 N

d) 445.156 kN 445.156*10^3 N

e) 405.670 kN 405.670*10^3 N

xiii)

a)

b)

c) 2,539.688 kN 2539.688*10^3 N

d) 2,539.688 kN 2539.688*10^3 N

e) 2,539.688 kN 2539.688*10^3 N

xiv)

AASHTO-LRFD-5.8.2.1:

a)below ;

b)

c)

d)

e)

f) N


On Outer Faces of Exterior Girder Vn= Vc-Ext-Outer. + Vs-Ext-Outer. Vn-Ext-Outer

On Inner Face of Exterior Girder, Vn = Vc-Ext-Gir-Inner + Vs-Ext-Gir-Inner Vn-Ext-Gir-Inner

On Position of Interior Girders, Vn = Vc-On-Inner+ Vs-On-Inner-Gir Vn-On-Inner-Gir

Computation of values for Nominal Shear Resistance (Vn) at Different Section of Abutment Cap under the

Provisions of AASHTO-LRFD-5.8.3.3 against Equation Vn = 0.25f/cbvdv + Vp (Equ. 5.8.3.3-2) :



On Outer Faces of Exterior Girder Vn= 0.25f/c-bv-Ext-Gir-Outer.dv-Ext-Gir-Outer. Vn-Ext-Gir-Outer.

On Inner Face of Exterior Girder, Vn= 0.25f/c-bv-Ext-Gir-Innerdv-Ext-Gir-Inner Vn-Ext-Gir-Inner

On Position of Interior Girders, Vn= 0.25f/c-bv-On-Inner-Girdv-On-Inner-Gir Vn-On-Inner-Gir

Chacking for requirment of Shear Reinforcement based on Computed Nominal Shear Resistance-Vn & the

Factored Shearing Resistance-Vr of Abutment Cap Accordting to the Provisions of AASHTO-LRFD-5.8.3.3 &




For RCC Component, Vp = 0.




g) 0.90

h)

& Equ. 5.8.3.3-2 are shown in the Table below :.

i) Table :- Checking for Requirement of Shear Reinforcement in respect of Computed values of Vn & Vr :

Location Calculated Relation Accepted Factored Relation

of Factored As per As per between Value of Shear between Shear Rein (SR)

Abutment Shear Equation. Equation Values of Resitance Values of Requires

Cap 5.8.3.3-1 5.8.3.3-2 otherwise SR

Section kN kN kN kN kN Not Require.

On Outer 465.719 445.156 2539.688 Vn-1< Vn-2 445.156 400.64 Vu> Vr Require SR

Face of

Ext.-Girder

On Inner 413.972 445.156 2539.688 Vn-1< Vn-2 445.156 400.64 Vu> Vr Require SR

Face of

Ext.-Girder

On 413.972 405.670 2,539.69 Vn-1< Vn-2 405.670 365.10 Vu> Vr Require SR

Int.-Girder

Position

xv) Checking of Required Max. Spacing for Transvers/Shear Reinforcement due to Applied Shearing Stress on Abutment Cap under provision of AASHTO-LRFD-5.8.2.7 :

a)

b)


d)


Segment. Value of Maximum Value of Relation Value of Value of Relation Relation Formula

Between Value of between between between Status

Abutment for the for the for the Whether

Cap Section Section Section Satisfy or

Sections mm mm mm Not Satisfy

End Edge 483.750 1.070 2.625 vu<0.1.25f/c 387.000 193.500 0.8dv<600 Satisfy

Ext.-Girder 0.4dv<300 Satisfy

Face


Checking for requirment of Shear Reinforcement in respect of Acceptable Nominal Shear Resistance, Vn, Respective

Factored Shear Resitance-Vr on Different Section of the Abutment Cap Computed under provision of Equ. 5.8.3.3-1

Vn-1 Vn-2 If VU > Vr

Vn

Force-Vu Vn-1 & Vn-2 Vr Vu& Vr

Vn-1 > Vn-2 Vu> Vr






c N/mm2


dv 0.125f/c 0.8dv 0.4dv

vu vu & 0.12f/c 0.8dv & 0.4dv &

sMax-1<600 sMax-2<300


Ext.-Girder 483.750 0.951 2.625 vu<0.1.25f/c 387.000 193.500 0.8dv<600 Satisfy

to 0.4dv<300 Satisfy

Int.-Girder

e) The Table indicates that all Sections of Abutment Cap have satisfied the required provisions for Spacings

f)is Minimum, thus it will Govern in Provision of same for Abutment Cap.

xvi) Computation of Spacing for Transverse/Shear Reinforcement for Abutment Cap :

a) 12.000 mmLength of Abutment Cap.

b) 226.195

c) Allowable Max. Spacing for Transverse/Shear Reinforcement 300.000 mm

d) The required Spacing for Transverse/Shear Reinforcement according to the 243.825 mm

f) Let Provide the 150mm Spacing for Transvers/Shear Reinforcements for total 150.000 mmLength of Abutment Cap.

xvii) Checking against Requirements of Minimum Transverse Reinforcements for Abument Cap :

a)

b) Table for Minimum Transverse/Shear Reinforcement for Abutment Cap.

Locations of Abutmen Web Width Provided Required Provided Status of Whether

Cap for Chacking for the Spacing of Minimum Shearing Provided & the Equation

Section Shear Rein Steel Area Steel Area Required have

Steel Area Satisfied or

mm mm Not Satiosfied

Ext. Girder OuterFace 1000.000 150.000 139.154 226.195 Av-pro> Av Satisfied

Ext. Girder OuterFace 1000.000 150.000 139.154 226.195 Av-pro> Av Satisfied

Position of Ent. Girder 1000.000 150.000 139.154 226.195 Av-pro> Av Satisfied

xviii)

a)

of Transverse/Shear Reinforcement under both the Equations, vu < 0.125f/c ; smax. = 0.8dv ≤ 600mm, (Equ. -

5.8.2.7-1). & vu > 0.125f/c ; smax. = 0.4dv ≤ 300mm, (Equ.-5.8.2.7-2).

Since under Equation 5.8.2.7-2 (smax. = 0.4dv ≤ 300mm) the Spacing of Transverse/Shear Reinforcement

Let Provide 2-Leged 12f Bars as Transverse/Shear Reinforcement for the Full DShear.


= 2*pDStir.2/4 mm2

sMax

sreq.

Provisions of AASHTO-LRFD-5.8.2.5;Equ.-5.8.2.5-1; Av ³ 0.083Öf/c(bvs/fy)

having s = Avfy/0.083Öf/cbv

s-pro.



bv spro Av-req. Av-pro

mm2 mm2 Av-pro>Av-req.

Checking for Width (b) of Abutment Cap in respect of Nominal Shearing Resistance (Vn) at Critical Section :

Against Applied Loads (DL & LL) the Critical Section for Shearing Forces Prevails at Inner Face of Exterior Girder.

b) 445.156 kN

445.156*10^3 N

c) 1,000.000 mm

e) 175.280 mm

Shear Depth for the Critical Section.

f)

xix) Checking in respect of Longitudinal Reinforcements (Tensile Reinforcements) Provided for Abutment Cap under Provision of AASHTO-LRFD-5.8.3.5 :


Where;

b) Variable

c) - MPa

d) -

e) - MPa

f) Variable N-mm


h) Variable mm

i) Variable N

j) Variable N

k) - N


Computed Nominal Shearing Resistance According to Provisions of AASHTO- Vn

LRFD--5.8.3.3 for the Section is Vn = 308.504*10^3N

Provided width of Abutment Cap, b = 1000 mm. bv-pro

In RCC Component the value of Vp = 0, thus according to Equ. 5.8.3.3-2, the bv-Cal.

Width of Abutment Cap, bv = Vn/0.25*f/cdv, where dv is Calculated Effective

Since the Calculated value of Girder width bv-Cal = bv-pro the Provided Girder Width, thus Design Critical Section is OK.


As is Area of Nonprestressing Steel on Flexural Tention side in mm2 As mm2
















m) 0.90

n) 0.90

o) 0.80 AASHTO-LRFD-5.5.4.2.

p)

q) Table- Showing Evalution of Equation-5.8.3.5-1 at Different Section of Abutment Cap & Status of Results.

Locations of Abutmen Calculted Calculted Status of

Cap for Chacking Provided Factored Factored Shearing Value of R/H Part Equation

Tensile Moment Shearing Resistance of the Wheter

Steel Area Force of Stirrups (L/H Part) Equation Satisfied or

kN-mm kN kN kN kN Not Satisfied

Ext. Girder OuterFace 3926.991 261.967 465.719 405.670 1610.066 914.13 Satisfied

Ext. Girder OuterFace 3926.991 165.589 413.972 405.670 1610.066 656.84 Satisfied

Position of Ent. Girder 3926.991 183.988 413.972 405.670 1610.066 691.07 Satisfied

In-between Girders 3926.991 206.986 0.000 405.670 1610.066 40.74 Satisfied

j) Since at all Section the requirments of Equation are being Satisfied, thus provision of Transverse/Shear Reinforcements for Abutment Cap is OK.

14 Shrinkage & Temperature Reinforcement on Vertical & Inclined Faces of Abutment Cap :

a) The Abutment Cap is being Flexural Designed having Longitudinal Reinforcements both on Top & Bottom Faces against Calculated respective Moments. It has also been Provided with Transverse/Shear Reinforcements against the Calculated Shearing Forces having their Position Perpendicular to Main Reinforcement both on Horizontal and Vertical Faces. Now it requires to Provide Horizontal Reinforcement on Vertical & Inclined Faces under Shrinkage

i) Shrinkage & Temperature Reinforcement on Vertica Faces ( Perpendicular to Traffic) :

a) Let consider 1 (One) meter Strip Length of Abutmebt Cap for Calculation of 1.000 mShrinkage & Temperature Reinforcement in Perpendicular to Traffic Direction 1,000.000 mmon its Vertical Faces.

b) 0.600 m 600.000 mm

c) 1.000 mm

c) 160.976

ff is Resistance Factor for Flexural Tension of Reinforced Concrete according ff

to AASHTO-LRFD-5.5.4.2.

fv is Resistance Factor for Shearing Force of Reinforced Concrete according fv

to AASHTO-LRFD-5.5.4.2,




As. Mu Vu Vs

Asfy

mm2


LStrip.

Depth of Vertical Face of Abutment Cap, hAb-Cap-Rec = 0.600 m hAb-Cap-Rec

Width of Abutment Cap, bAb-Cap = 1.000m bAb-Cap

According to AASHTO-LRFD-5.10.8.1. Compoent having Less Than As-req-T&S-VF mm2

Shrinkage & Temperature Reinforcement; Here,

d) 600000.000

e) 16 mmFaces of Girder.

f) 201.062

g) 749.413 mm,C/C

h)

i) 3,000.000 mm

j) Value of Max. Spacing = 450mm 450.000 mm

k) Allowable Maximum Spacing for Shrinkage & Temperature Reinforcements 450.000 mm

h) 2.222 nos.

j) 4 nosFaces (Earth, Water & Sides) of Abutment Cap.

k) 200.000 mm

l) 804.248 & Temperature Reinforcement on Vertical Surfaces of Abutment Cap

m)

m)

15 Checking Against Punhing Shear Force on Abutment Cap due to Vertical Loads from Superstruture (DL& LL) & Also Loads from Abutment Cap & Back Wall :

a)Section should be considered under the Provision of Article 5.8.3.2. According to which the Critacal Section is at a

1200mm Thicness will require a Steel Area, As ³ 0.11Ag/fy as

Ag is Gross Area of Girder's Each Vertical Face = LStrip*hAb-Cap-Rec Ag-Hori. mm2



Spacing of Reinforcement with 16f bars = Af16.b/As-req-S&T.Ab-Cap. sreq-1

According to AASHTO-LRFD-5.10.8.2 the Max. Spacing of Shrinkage & Temperature Reinforcements should be

the Less vale of 3-times of Component thickness = 3*bAb-Cap or = 450 mm.

Value of Max. Spacing with 3 Times of Component Thickness = 3*bAb-Cap sreq-2

sreq-3

sallow.

According to AASHTO-LRFD-5.10.8.2 Vertical Faces of Abutment Cap

Number of 16f bars required against Allowable Max. Spacing = hAb-Cap/sallow NBar-req

Let Provide 4 nos. 16f Bars as Shrinkage & Temperature on all Vertical NBar-pro-VF

Spacing for provided 4 nos.16f bars as Longitudinal Shrinkage & spro-S&T-VF

Temperature on Vertical Faces = hAb-Cap/(NBar-pro-VF. - 1)

Steel Area against Provided 4 nos. 16f bars as Horizontal Shrinkage As-pro-S&T-VF mm2

= Af-16*NBar-pro-VF.

Based on above Calculations Let Provide on Inclined Water Face 1 (One) no. 16f Bar & on Inclined Earth Face2 (Two) nos. 16f Bars as Horizontal Shrinkage & Temperature Reinforcement.

Since Provided S&T Steel Area As-pro-S&T-VF. > As-req-S&T-VF. & Provided Spacing of S&T spro-S&T-VF..> s-req-S&T-VF. thus the provision of Horizontal Shrinkage & Temperature Reinforcement on Vertical Faces of Abutment Cap onSurfaces is Faces are OK.

The Abutment Cap is affect by One Way Action of Loading, and According to AASHTO-LRFD-5.13.3.6 its Critical

b)Earthen Edge from Stem Face = 400mm, which less than dv, thus it is not required to any Checking in respect ofPunching Shear Force on Abutment Cap. More over the Abutment Cap is being Provided with Transverse & ShearReinforcements both on Horizontal (Perpendicular to Abutment Cap on Top & Bottom) & Vertical Faces against the Calculated Factored Shearing Forces, which will meet the requirements if any.

Distance dv (The Effective Shear Depth of Abutment Cap) from the Earthen Face of Abutment Stem.

Since the Calculated Effective Shear Depth of Abutment Cap, dv = 483.750 mm and the Distance of Abutment Cap

L.

Part-I :- Structural Dimensions, Multiplying Factors & Computation of Different Load Intensities.

1 Sketch Diagram of Abutment Wall & Wing Walls :

2 Structural Type, Design Criteria & Dimentions of Structure :

i) Type of Abutment : Wall Type Abutment.

ii) Type of Wing-walls : Wall Type Wing Walls Integrated with Abutment Wall having Counterforts over Well & Cantilever Wings beyond Well.

iii) Design Criteria : Strength Limit State (USD) Design According to AASHTO-LRFD-2004.

iv) Dimensions of Substructural Components :




Structural Design of Abutment Wall & Wing Walls for Aboutments :

C

C

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

5225

1200

2000

6350

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

10250

9350

12750

3450 3450 3450600 600 600600

34503450

60 0

2750

60 0

2525

1775

5500

3000

450

450

450

450

2150

2150

2450

2750

RL-5.00m

RL-2.20m

3000 2100

300



e) Height of Back Wall 2.147 m

f) Height of Wing Wall 4.947 m

g) Length of Wing Walls upon Well cap 2.975 m

h) Width of Abutment Well Cap (Longitudinal Length), 5.500 m

i) Length of Abutment Well Cap (Transverse Length), 12.750 m

j) 10.250 mDirection.

k) Inner Distance in between Wing Walls (Transverse), 9.350 m

l) Thickness of Abutment Wall (Stem) at Bottom 0.750 m

m) Thickness of Abutment Wall (Stem) at Top 0.450 m

n) Thickness of Counterfort Wall (For Wing Wall) 0.450 m

o) Number of Wing-Wall Counterforts (on each side) 1.000 No's

p) Clear Spacing between Counerfort & Abutment Wall at Bottom 1.775 m

q) Average Spacing between Counerfort & Abutment Wall 2.375 m

r) 2.825 m

s) Thickness of Wing Walls within Well Cap, 0.450 m

t) Thickness of Cantilever Wing Walls 0.450 m

u) Length of Cantilever Wing Walls 3.000 m

v) Height of Rectangular Portion of Cantilever Wing Walls 2.000 m

w) Height of Triangular Portion of Cantilever Wing Walls 1.500 m

x) Longitudinal Length of Well Cap on Toe Side from Abutment Wall 2.525 mOuter Face.

y) Average Length (Longitudinal) of Well Cap on Heel Side from 2.825 m

hWell-Cap.

hSteam.

hb-wall

H-W-Wall

LW-W-Well-Cap

WAb-Cap

LAb-T-W-Cap


LAb-T-Inner

t.-Ab-wall-Bot.

t.-Ab-wall-Top.

tWW-Countf.

NW-W-count


SAver-Count&Ab.



tWW

tCant-WW

Lw-wall-Cant.

hw-wall-Cant.-Rec.

hw-wall-Cant.-Tri.

L-W-Cap-Toe.

L-W-Cap-Heel-Aver.

3 Design Data in Respect of Unit Weight, Strength of Materials & Load Multiplier Factors :

i)

9.807






ii)







a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.89

e) 2.89 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200,000.000 MPa


a) 8.384 n 8

b) r 20c) k 0.29 d) j 0.90

e) R 1.10






gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


c



c Ec













fFlx-Rin.









j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85



of 90 km/hr) for normal wind load only are selected as CRITICAL conditions for bridge structure.









ii) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.



but those would be occasional. Thus the respective Multiplier Factors of Limit State STRENGTH I (Bridge used by Normal Vehicle without wind load) for normal operation, Limit State of STRENGTH-III (Wind Velocity exceeding90km/hr) for wind load during cyclonic storm condition and Limit State of STRENGTH-IV (Having Wind Velocity



gEH


gEV


gES


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000






q) -

r) -

t) 1.000


i) Permanent & Dead Load Multiplier Factors for Service Limit State Design (WSD) According to AASHTO-




















LRFD-3.4.1 ; Table 3.4.1-1&2 :










b) 1.000

c) IM 1.300 AASHTO LRFD-3.6.2.1, Table 3.6.2.1-1; SERVICE - II(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000



gEH


gEV


gES


















q) -

r) -

t) 1.000



a) 0.441

b) f 34.000



d) 0.34 to 0.45 dim



0.008


c) 0.018

18.000













O


Dp-ES kN/m2








0.004761

4.761



0.004761

4.761

c) k 0.441

d) 1,835.424

e) g 9.807

f) 900.000 mm




h) 1,050.000 mm



600.000 mm 0.600 m

Part-II :- Computation of Different Load Events, Respective Moments & Shear.

1 Philosophy in Flexural Design of Abutment Walls & Wing Walls of Bridge Structure :

a)

reduce the affect of Horizontal Pressure a great extent. But in Construction Phenomenon execution Abutment Wall works are to be done a well ahead before Construction of Super structural Components having the total Horizontal


kN/m2

Dp-LS = kgEgheq*10-9 Dp-LL-Ab³6.00m N/mm2

kN/m2


kN/m2


kN/m2






Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.

Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

The Abutment Walls of Bridge Substructure will have to face both Vertical Dead Loads (DL) & Live Loads (LL) from the Superstructure and Horizontal Dead Load (DL) & Live Load (LL) Pressures due to Earth & Surcharge on their Back Faces. Under Vertical Loads (DL & LL) Abutment Wall will behave as Compression Component, whereas it will be a Cantilever one under Horizontal Pressure (DL & LL). The action of Vertical Loads from the Superstructure will

Loads (DL & LL) upon it. Thus to ensure the sustainability of Abutment Wall against the affect of Horizontal Loads

b)

c) 1.000 mForces & Moments both Abutment & Wing Wall it is considered having 1.000m Length & Wide Strips according to sequences of Design.

2

i) Horizontal Loads at Bottom Level of Abutment Wall on 1 (One) meter Strip :

a) 39.252

b) 7.935

c) 7.141

d) 47.187

ii) Horizontal Loads at Bottom Level of Wing Walls on 1 (One) meter Strip :

a) 39.252

b) 7.935

c) 8.331

d) 47.187

3

i) Factored Horizontal Loads on Abutment Wall for 1 (One) meter Wide Strip :

a) 58.878

& Forces (DL & LL) it should be Designed as a Cantilever Flexural Component.

The Wing Walls will not have any Vertical Loads but those will have to face similar Horizontal Dead Load (DL) & Live Load (LL) Pressures caused by Earth & Surcharge Loads. Thus Wing Walls are also have to Designed as Cantilever Flexural Components against Horizontal Loads & Forces (DL & LL) upon those.

For the Design purpose & Calculation of Imposed Loads (DL & LL), Shearing LH&V

Computation of Horizontal Loads on Abutment & Wing Walls due to Lateral Soil & Surcharge Pressure :

Horizontal Pressure Intensity due to Lateral Soil Dead Load (DL) at PDL-H-Soil-Ab kN/m2/m

Bottom Level of Abutment Wall (Perpendicular to Traffic). = k0*wE*H1

Horizontal Pressure Intensity due to Surcharge Dead Load (DL) on PDL-H-Sur-Ab kN/m2/m

Abutmen Wall (Perpendicular to Traffic). = Dp-ES.LH&V

Horizontal Pressure Intensity due to Surcharge Live Load (LL) on PLL-H-Sur-Ab kN/m2/m

Abutmen Wall (Perpendicular to Traffic) having H1 < 6.000m. = Dp-LL-Ab.LH&V

Total Dead Load (DL) Horizontal Pressure Intensity on Abutment Wall pDL-H-Total-Ab kN/m2/m

= PDL-H-Soil-WW + PDL-H-Sur-WW

Horizontal Pressure Intensity due to Lateral Soil Dead Load (DL) PDL-H-Soil-WW kN/m2/m

at Bottom Level of Wing Walls (Parallel to Traffic). = k0*wEH1

Horizontal Pressure Intensity due to Surcharge Dead Load (DL) on PDL-H-Sur-WW kN/m2/m

Wing Walls (Paralle to Traffic). = Dp-ES.LH&V

Horizontal Pressure Intensity due to Surcharge Live Load (LL) on pLL-H-Sur-WW kN/m2/m

Wing Wall (Parallel to Traffic) having H1 < 6.000m. = Dp-LL-WW.LH&V

Total Dead Load (DL) Horizontal Pressure Intensity on Wing Wall pDL-H-Total-WW kN/m2/m

= PDL-H-Soil-WW + PDL-H-Sur-WW

Factored Horizontal Loads on Abutment & Wing Walls under Strength Limit State (USD):

Factored Horizontal Pressure Intensity due to Lateral Soil Dead FPDL-H-Soil-Ab-USD kN/m2/m

Load (DL) at Bottom Level of Abutment Wall (Perpendicular to Traffic). = gEH*PDL-H-Soil-Ab

b) Factored Horizontal Pressure Intensity due to Surcharge Dead 11.902

c) Factored Horizontal Pressure Intensity due to Surcharge Live 12.496


ii) Factored Horizontal Loads at Bottom Level of Wing Walls on 1 (One) meter Wide Strip :

a) Factored Horizontal Pressure Intensity due to Lateral Soil Dead 58.878



d) 85.359

4

i) Factored Horizontal Loads on Abutment Wall for 1 (One) meter Wide Strip :

a) 39.252




ii) Factored Horizontal Loads at Bottom Level of Wing Walls on 1 (One) meter Wide Strip :

a) 39.252


FPDL-H-Sur-Ab-USD kN/m2/m

Load (DL)on Abutment Wall (Perpendicular to Traffic). = gES*PDL-H-Sur-Ab

FPLL-H-Sur-Ab-USD kN/m2/m

Load (LL) on Abutment Wall (Perpendicular to Traffic). = mgLL-LS*PLL-H-Sur-Ab

åFPH-Ab-USD kN/m2/m(DL & LL) at Bottom Level of Abutment (Perpendicular to Traffic).

FPDL-H-Soil-WW-USD kN/m2/m

Load (DL) at Bottom Level of Wing Walls (Paralle to Traffic). = gEH*PDL-H-Soil-WW

FPDL-H-Sur-WW-USD kN/m2/m

Load (DL) on Wing Walls (Paralle to Traffic). = gES*PDL-H-Sur-WW

FPLL-H-Sur-WW-USD kN/m2/m

Load (LL) on Wing Walls (Paralle to Traffic). = mgLL-LS*PLL-H-Sur-WW

Total Factored Horizontal Pressure Intensity due to All Applied Loads åFPH-WW-USD kN/m2/m(DL & LL) at Bottom Level of Wing Walls (Paralle to Traffic)..

Factored Horizontal Loads on Abutment & Wing Walls under Service Limit State (WSD):

Factored Horizontal Pressure Intensity due to Lateral Soil Dead FPDL-H-Soil-Ab-WSD kN/m2/m

Load (DL) at Bottom Level of Abutment Wall (Perpendicular to Traffic). = gEH*PDL-H-Soil-Ab

FPDL-H-Sur-Ab-WSD kN/m2/m

Load (DL)on Abutment Wall (Perpendicular to Traffic). = gES*PDL-H-Sur-Ab

FPLL-H-Sur-Ab-WSD kN/m2/m

Load (LL) on Abutment Wall (Perpendicular to Traffic). = mgLL-LS*PLL-H-Sur-Ab

åFPH-Ab-WSD kN/m2/m(DL & LL) at Bottom Level of Abutment (Perpendicular to Traffic).

Factored Horizontal Pressure Intensity due to Lateral Soil Dead FPDL-H-Soil-WW-WSD kN/m2/m

Load (DL) at Bottom Level of Wing Walls (Paralle to Traffic). = gEH*PDL-H-Soil-WW

FPDL-H-Sur-WW-WSD kN/m2/m

Load (DL) on Wing Walls (Paralle to Traffic). = gES*PDL-H-Sur-WW


d) 55.518

5 Computation of Coefficients for Calculation of Moments on Abutment Wall & Wing Walls:

i) Span Ratio of Abutment & Wing Walls in respect of Horizontal to Vertical Spans :

a) Horizontal Span Length of Abutment Wall (Inner Distance) 9.350 m

b) Horizontal Span of Length of Wing Walls (Inner Distance between 2.375 mAbutment Wall & Wing Wall Counterfort).

c) Total Vertical Depth of Abutment Wall/Wing Walls (From Back Wall Top upto H1 4.947 mWell Cap Top)

d) Abutment Span Ratio for Horizontal to Vertical with Freely Supported on 1.890

e) Wing Wall Span Ratio for Horizontal to Vertical with Un-supported on 0.480

g) Abutment Span Ratio for Vertical to Horizontal with Freely Supported on 0.529

h) Wing Wall Span Ratio for Vertical to Horizontal with Un-supported on 2.083

ii) Coefficients for Calculation of Moments for Freely Supported Triangular Loadings on Abutment Walls :(Using Table-53 of Reinforced Concrete Design Handbook UK.)

a) k 1.890

b) (+) ve Moment Coefficient of Abutment Wall for Vertical Span Strip 0.033

c) 0.005

d) 0.055 Vertical Span Strip at Middle Position under Triangular Loading

e) (-) ve Moment Coefficient of Abutment Wall for Horizontal Span Strip 0.003 on Face of Vertical Edge

iii) Coefficients for Calculation of Moments for Unsupported Triangular Loadings on Wing Walls :(Using Table-53 of Reinforced Concrete Design Handbook UK.)

FPLL-H-Sur-WW-WSD kN/m2/m

Load (LL) on Wing Walls (Paralle to Traffic). = mgLL-LS*PLL-H-Sur-WW

Total Factored Horizontal Pressure Intensity due to All Applied Loads åFPH-WW-WSD kN/m2/m(DL & LL) at Bottom Level of Wing Walls (Paralle to Traffic)..

LAb-T-Inner

SAver-Count&Ab.

kAb-Supp.-H/V

Top = LAb-T-Inner/H1

kWW-Un-Supp.-H/V

Top = SAver-Count&Ab./H1.

kAb-Supp.-V/H

Top = H1/LAb-T-Inner

kWW-Un-Supp.-V/H

Top = H1/SAver-Count&Ab.

k = 1.890

(+)cS-V-Ab-T

(+) ve Moment Coefficient of Abutment Wall for Horizontal Span Strip (+)cS-H-Ab-T

(-) ve Moment Coefficient on Bottom Surface of Abutment Wall for (-)cS-V-Ab-T

(-)cS-H-Ab-T

a) k 0.480

b) (+) ve Moment Coefficient of Wing Wall for Vertical Span Strip. -

c) (+) ve Moment Coefficient of Wing Wall for Horizontal Span Strip. 0.038

0.009 d) Position on Bottom Surface.

e) (-) ve Moment Coefficient at of Wing Wall for Horizontal Span Strip 0.058 on Face of Vertical Edge

iv) Coefficients for Calculation of Moments related to Uniformly Loadings on Abutment Walls :(Using Table-4.2; 4.3 & 4.4 of Book Design of Concrete Structures- G.Winter. Seventh Edition)

a) k 0.550

b) 0.052 Strip.

c) 0.005 Span Strip.

d) 0.070 Strip.

e) 0.007 Span Strip.

f) 0.085 Abutment Vertical Span Strip

g) 0.014 Abutment Horizontal Span Strip

v) Coefficients for Calculation of Moments related to Uniformly Loadings on Wing Walls :(Using Table-4.2; 4.3 & 4.4 of Book Design of Concrete Structures- G.Winter. Seventh Edition)

a) k 1.000




k = 0.480

(+)cS-V-WW-T

(+)cS-H-WW-T

(-) ve Moment Coefficient of Wing Wall for Vertical Span Strip at Middle (-)cS-V-WW-T

(-)cS-H-WW-T

Since k = 0.529 > 0.500, Let Consider k = 0.550

(+) ve Moment Coefficient for Dead Load (DL) on Abutment Vertical Span (+)cS-V-Ab-U-DL

(+) ve Moment Coefficient for Dead Load (DL) on Abutment Horizontal (+)cS-H-Ab-U-DL

(+) ve Moment Coefficient for Live Load (LL) on Abutment Vertical Span (+)cS-V-Ab-U-LL

(+) ve Moment Coefficient for Live Load (LL) on Abutment Horizontal (+)cS-H-Ab-U-LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-V-Ab-U-DL+LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-H-Ab-U-DL+LL

Since k > 2.083, Let Consider k = 1.000.

(+) ve Moment Coefficient for Dead Load (DL) on Wing Wall Vertical (+)cS-V-WW-U-DL

(+) ve Moment Coefficient for Dead Load (DL) on Wing Wall Horizontal (+)cS-H-WW-U-DL

(+) ve Moment Coefficient for Live Load (LL) on Wing Wall Vertical (+)cS-V-WW-U-LL


g) 0.033 Wing Wall Vertical Span Strip

h) 0.061 Wing Wall Horizontal Span Strip

6 Factored Moments due to Horizontal Loads on Abutment under Strength Limit State (USD):

i)

a) (+) ve Moment on Vertical Span due to Traingular Soil Load 47.550 kN-m/m


c) (+) ve Moment on Vertical Span due to Uniform Surcharge Live 21.407 kN-m/m


ii)


b) (-) ve Moment on Vertical Span due to Uniform Surcharge Dead 24.758 kN-m/m

c) (-) ve Moment on Vertical Span due to Uniform Surcharge Live 25.995 kN-m/m


iii)

a) (+) ve Moment on Horizontal Span due to Traingular Soil Load 25.736 kN-m/m



(+) ve Moment Coefficient for Live Load (LL) on Wing Wall Horizontal (+)cS-H-WW-U-LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-V-WW-U-DL+LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-H-WW-U-DL+LL

Calculation of Factored (+) ve Moments on Abutment Wall in Vertical Span using Coefficients :

(+)MS-V-Ab-T-USD

= (+)cS-V-Ab-T*FPDL-H-Soil-Ab-USD*H12

(+)MS-V-Ab-Sur-DL-USD

Load = (+)cS-V-Ab-U-DL*FPDL-H-Sur-Ab-USD*H12

(+)MS-V-Ab-Sur-LL-USD

Load = (+)cS-V-Ab-U-LL*FPLL-H-Sur-Ab-USD*H12

(+)åMS-V-Ab-USD

Calculation of Factored (-) ve Moments on Abutment Wall in Vertical Span using Coefficients :

(-)MS-V-Ab-T-USD

= (-)cS-V-Ab-T*FPDL-H-Soil-Ab-USD*H12

(-)MS-V-Ab-Sur-DL-USD

Load = (-)cS-V-Ab-U-DL+LL*FPDL-H-Sur-Ab-USD*H12

(-)MS-V-Ab-Sur-LL-USD

= (-)cS-V-Ab-U-DL-LL*FPLL-H-Sur-Ab-USD*H12

(-)åMS-V-Ab-USD

Calculation of Factored (+) ve Moments on Abutment Wall in Horizontal Span using Coefficients :

(+)MS-H-Ab-T-USD

= (+)cS-H-Ab-T*FPDL-H-Soil-Ab-USD*LAb-T-Inner2

(+)MS-H-Ab-Sur-DL-USD

Load = (+)cS-H-Ab-U-DL*FPDL-H-Sur-Ab-USD*LAb-T-Inne2

(+)MS-H-Ab-Sur-LL-USD

Load = (+)cS-H-Ab-U-LL*FPLL-H-Sur-Ab-USD*LAb-T-Inne2


iv)

a) (-) ve Moment on Horizontal Span due to Traingular Soil Load 10.295 kN-m/m




7 Factored Moments due to Horizontal Loads on Wing Walls under Strength Limit State (USD):

i)





ii)





iii)


(+)åMS-H-Ab-USD

Calculation of Factored (-) ve Moments on Abutment Wall in Horizontal Span using Coefficients :

(-)MS-H-Ab-T-USD

= (-)cS-H-Ab-T*FPDL-H-Soil-Ab-USD*LAb-T-Inne2

(-)MS-H-Ab-Sur-DL-USD

Load = (-)cS-H-Ab-U-DL+LL*FPDL-H-Sur-Ab-USD*LAb-T-Inne2

(-)MS-H-Ab-Sur-LL-USD

Load = (-)cS-H-Ab-U-DL-LL*FPLL-H-Sur-Ab-USD*LAb-T-Inne2

(-)åMS-H-Ab-USD

Calculation of Factored (+) ve Moments on Wing Wall in Vertical Span using Coefficients :

(+)MS-V-WW-T-USD

= (+)cS-V-WW-T*FPDL-H-Soil-WW-USD*H12

(+)MS-V-WW-Sur-DL-USD

Load = (+)cS-V-WW-U-DL*FPDL-H-Sur-WW-USD*H12

(+)MS-V-WW-Sur-LL-USD

Load = (+)cS-V-WW-U-LL*FPLL-H-Sur-WW-USD*H12

(+)åMS-V-WW-USD

Calculation of Factored (-) ve Moments on Wing Wall in Vertical Span using Coefficients :

(-)MS-V-WW-T-USD

= (-)cS-V-WW-T*FPDL-H-Soil-WW-USD*H12

(-)MS-V-WW-Sur-DL-USD

Load = (-)cS-V-WW-U-DL+LL*FPDL-H-Sur-WW-USD*H12

(-)MS-V-WW-Sur-LL-USD

= (-)cS-V-WW-U-DL-LL*FPLL-H-Sur-WW-USD*H12

(-)åMS-V-WW-USD

Calculation of Factored (+) ve Moments on Wing Wall in Horizontal Span using Coefficients :

(+)MS-H-WW-T-USD




iv)





8 Factored Moments due to Horizontal Loads on Abutment under Service Limit State (WSD):

i)





ii)




= (+)cS-H-WW-T*FPDL-H-Soil-WW-USD*SAver-Count&Ab2

(+)MS-H-WW-Sur-DL-USD

Load = (+)cS-H-WW-U-DL*FPDL-H-Sur-Ab-USD*SAver-Count&Ab2

(+)MS-H-WW-Sur-LL-USD

Load = (+)cS-H-WW-U-LL*FPLL-H-Sur-WW-USD*SAver-Count&Ab2

(+)åMS-H-WW-USD

Calculation of Factored (-) ve Moments on Wing Wall in Horizontal Span using Coefficients :

(-)MS-H-WW-T-USD

= (-)cS-H-WW-T*FPDL-H-Soil-WW-USD*SAver-Count&Ab2

(-)MS-H-WW-Sur-DL-USD

Load = (-)cS-H-WW-U-DL+LL*FPDL-H-Sur-WW-USD*SAver-Count&Ab2

(-)MS-H-WW-Sur-LL-USD

Load = (-)cS-H-WW-U-DL-LL*FPLL-H-Sur-WW-USD*SAver-Count&Ab2

(-)åMS-H-WW-USD

Calculation of Factored (+) ve Moments on Abutment Wall in Vertical Span using Coefficients :

(+)MS-V-Ab-T-WSD

= (+)cS-V-Ab-T*FPDL-H-Soil-Ab-WSD*H12

(+)MS-V-Ab-Sur-DL-WSD

Load = (+)cS-V-Ab-U-DL*FPDL-H-Sur-Ab-WSD*H12

(+)MS-V-Ab-Sur-LL-WSD

Load = (+)cS-V-Ab-U-LL*FPLL-H-Sur-Ab-WSD*H12

(+)åMS-V-Ab-WSD

Calculation of Factored (-) ve Moments on Abutment Wall in Vertical Span using Coefficients :

(-)MS-V-Ab-T-WSD

= (-)cS-V-Ab-T*FPDL-H-Soil-Ab-WSD*H12

(-)MS-V-Ab-Sur-DL-WSD

Load = (-)cS-V-Ab-U-DL+LL*FPDL-H-Sur-Ab-WSD*H12

(-)MS-V-Ab-Sur-LL-WSD


iii)





iv)





9 Factored Moments due to Horizontal Loads on Wing Walls under Service Limit State (WSD):

i)





ii)

= (-)cS-V-Ab-U-DL-LL*FPLL-H-Sur-Ab-WSD*H12

(-)åMS-V-Ab-WSD

Calculation of Factored (+) ve Moments on Abutment Wall in Horizontal Span using Coefficients :

(+)MS-H-Ab-T-WSD

= (+)cS-H-Ab-T*FPDL-H-Soil-Ab-WSD*LAb-T-Inner2

(+)MS-H-Ab-Sur-DL-WSD

Load = (+)cS-H-Ab-U-DL*FPDL-H-Sur-Ab-WSD*LAb-T-Inne2

(+)MS-H-Ab-Sur-LL-WSD

Load = (+)cS-H-Ab-U-LL*FPLL-H-Sur-Ab-USD*LAb-T-Inne2

(+)åMS-H-Ab-WSD

Calculation of Factored (-) ve Moments on Abutment Wall in Horizontal Span using Coefficients :

(-)MS-H-Ab-T-WSD

= (-)cS-H-Ab-T*FPDL-H-Soil-Ab-WSD*LAb-T-Inne2

(-)MS-H-Ab-Sur-DL-WSD

Load = (-)cS-H-Ab-U-DL+LL*FPDL-H-Sur-Ab-WSD*LAb-T-Inne2

(-)MS-H-Ab-Sur-LL-USD

Load = (-)cS-H-Ab-U-DL-LL*FPLL-H-Sur-Ab-USD*LAb-T-Inne2

(-)åMS-H-Ab-WSD

Calculation of Factored (+) ve Moments on Wing Wall in Vertical Span using Coefficients :

(+)MS-V-WW-T-WSD

= (+)cS-V-WW-T*FPDL-H-Soil-WW-WSD*H12

(+)MS-V-WW-Sur-DL-WSD

Load = (+)cS-V-WW-U-DL*FPDL-H-Sur-WW-WSD*H12

(+)MS-V-WW-Sur-LL-WSD

Load = (+)cS-V-WW-U-LL*FPLL-H-Sur-WW-WSD*H12

(+)åMS-V-WW-WSD

Calculation of Factored (-) ve Moments on Wing Wall in Vertical Span using Coefficients :





iii)





iv)





10 Unfactored Moments due to Horizontal Dead Loads on Abutment Wall :

i)



(-)MS-V-WW-T-WSD

= (-)cS-V-WW-T*FPDL-H-Soil-WW-WSD*H12

(-)MS-V-WW-Sur-DL-WSD

Load = (-)cS-V-WW-U-DL+LL*FPDL-H-Sur-WW-USD*H12

(-)MS-V-WW-Sur-LL-WSD

Load = (-)cS-V-WW-U-DL-LL*FPLL-H-Sur-WW-WSD*H12

(-)åMS-V-WW-WSD

Calculation of Factored (+) ve Moments on Wing Wall in Horizontal Span using Coefficients :

(+)MS-H-WW-T-WSD

= (+)cS-H-WW-T*FPDL-H-Soil-WW-WSD*SAver-Count&Ab2

(+)MS-H-WW-Sur-DL-WSD

Load = (+)cS-H-WW-U-DL*FPDL-H-Sur-Ab-WSD*SAver-Count&Ab2

(+)MS-H-WW-Sur-LL-WSD

Load = (+)cS-H-WW-U-LL*FPLL-H-Sur-WW-WSD*SAver-Count&Ab2

(+)åMS-H-WW-WSD

Calculation of Factored (-) ve Moments on Wing Wall in Horizontal Span using Coefficients :

(-)MS-H-WW-T-WSD

= (-)cS-H-WW-T*FPDL-H-Soil-WW-WSD*SAver-Count&Ab2

(-)MS-H-WW-Sur-DL-USD

Load = (-)cS-H-WW-U-DL+LL*FPDL-H-Sur-WW-USD*SAver-Count&Ab2

(-)MS-H-WW-Sur-LL-WSD

Load = (-)cS-H-WW-U-DL-LL*FPLL-H-Sur-WW-WSD*SAver-Count&Ab2

(-)åMS-H-WW-WSD

Calculation of (+) ve Moments on Abutment Wall in Vertical Span due to Unfactored Dead Loads :

(+)MS-V-Ab-T-UF

= (+)cS-V-Ab-T*PDL-H-Soil-Ab*H12

(+)MS-V-Ab-Sur-DL-UF

Load = (+)cS-V-Ab-U-DL*PDL-H-Sur-Ab*H12


ii)




iii)




iv)



c) Total (-) ve Moment on Horizontal Span Strip for all Applied Dead Loads 20.006 kN-m/m

11 Unfactored Moments due to Horizontal Dead Loads on Wing Walls :

i)




ii)


(+)åMUF-S-V-Ab-UF

Calculation of (-) ve Moments on Abutment Wall in Vertical Span due to Unfactored dead Loads :

(-)MS-V-Ab-T-UF

= (-)cS-V-Ab-T*PDL-H-Soil-Ab*H12

(-)MS-V-Ab-Sur-DL-UF

= (-)cS-V-Ab-U-DL+LL*PDL-H-Sur-Ab*H12

(-)åMS-V-Ab-UF

Calculation of (+) ve Moments on Abutment Wall in Horizontal Span due to Unfactored Dead Loads :

(+)MS-H-Ab-T-UF

= (+)cS-H-Ab-T*PDL-H-Soil-Ab*LAb-T-Inner2

(+)MS-H-Ab-Sur-DL-UF

Load = (+)cS-H-Ab-U-DL*PDL-H-Sur-Ab*LAb-T-Inne2

(+)åMS-H-Ab-UF

Calculation of (-) ve Moments on Abutment Wall in Horizontal Span due to Unfactored Dead Loads :

(-)MS-H-Ab-T-UF

= (-)cS-H-Ab-T*PDL-H-Soil-Ab*LAb-T-Inne2

(-)MS-H-Ab-Sur-DL-UF

Load = (-)cS-H-Ab-U-DL+LL*PDL-H-Sur-Ab*LAb-T-Inne2

(-)åMS-H-Ab-UF

Calculation of (+) ve Moments on Wing Wall in Vertical Span due to Unfactored Dead Loads :

(+)MS-V-WW-T-UF

= (+)cS-V-WW-T*PDL-H-Soil-WW*H12

(+)MS-V-WW-Sur-DL-UF

Load = (+)cS-V-WW-U-DL*PDL-H-Sur-WW*H12

(+)åMS-V-WW-UF

Calculation of (-) ve Moments on Wing Wall in Vertical Span due to Unfactored Dead Loads :

(-)MS-V-WW-T-UF



iii)




iv)



c) Total (-) ve Moment on Horizontal Span Strip for all Applied Loads 15.572 kN-m/m

12 Computation of Factored Horizontal Shearing Forces on Abutment & Wing Walls (USD) :

i) Coefficients for Calculation of Shearing Froces on Abutment & Wing Walls Vertical & Horizontal Faces :(Using Table-4.5 of Book Design of Concrete Structures- G.Winter. Seveth Edition.)

a) k 0.550

b) k 1.000

c) Coefficiant of Shearing Forces for Vetical Span Strip having Shearing Forces 0.850 Bottom Edge of Abutment Wall

d) Coefficiant of Shearing Forces for Horizontal Span Strip having Shearing 0.150 Forces on Abutment Wall Vertical Faces

e) Coefficiant of Shearing Forces for Vetical Span Strip having Shearing Forces 0.330 Bottom Edge of Wing Wall

f) Coefficiant of Shearing Forces for Horizontal Span Strip having Shearing 0.670 Forces on Wing Wall Vertical Faces.

= (-)cS-V-WW-T*PDL-H-Soil-WW*H12

(-)MS-V-WW-Sur-DL-UF

= (-)cS-V-WW-U-DL+LL*PDL-H-Sur-WW*H12

(-)åMS-V-Ab-UF

Calculation of (+) ve Moments on Wing Wall in Horizontal Span due to Unfactored Dead Loads :

(+)MS-H-WW-T-UF

= (+)cS-H-WW-T*PDL-H-Soil-WW*SAver-Count&Ab2

(+)MS-H-WW-Sur-DL-UF

Load = (+)cS-H-WW-U-DL*PDL-H-Sur-WW*SAver-Count&Ab2

(+)åMS-H-WW-UF

Calculation of (-) ve Moments on Wing Wall in Horizontal Span due to Unfactored Dead Loads :

(-)MS-H-WW-T-UF

= (-)cS-H-WW-T*PDL-H-Soil-WW*SAver-Count&Ab2

(-)MS-H-WW-Sur-DL-UF

Load = (-)cS-H-WW-U-DL+LL*PDL-H-Sur-WW*SAver-Count&Ab2

(-)åMS-H-WW-UF

Since for Abutment k = 0.529 > 0.500, Let Consider k = 0.550

Since Wing Walls k > 2.083, Let Consider k = 1.000.

cShear-V-Ab

cShear-H-Ab

cShear-V-WW

cShear-H-WW

ii) Calculation of Shearing Froces on Abutment & Wing Walls Vertical & Horizontal Faces :

a) Total Factored Horizontal Load on Aburment Wall for 1.000m Wide 266.333 kN/m

b) Total Factored Horizontal Load on Wing Wall for 1.000m Wide Strip 276.636 kN/m

c) Shearing Forces on Abutment Wall at its Bottom Level for Vertical Span 226.383 kN/m

d) Shearing Forces on Abutment Wall at its Bottom Level for Horizontal Span 39.950 kN/m

c) Shearing Forces on Wing Wall at its Bottom Level for Vertical Span Strip 91.290 kN/m

d) Shearing Forces on Wing Wall at its Bottom Level for Horizontal Span 185.346 kN/m

Part-III :- Flexural Design of Abutment Wall Against Applied Loads Under Strength Limit State (USD).

1 Computation of Related Features required for Flexural Design of Vertical & Horizontal Reinforcements onAbutment Wall :

i) Design Strip Width for Abutment Wall in Vertical & Horizontal Direction & Clear Cover on different Faces;

a) b 1.000 m = 1000mm

b) 75.000 mm

50.000 mm



b) c Variable

c) Variable

Variable

Variable

410.000

Variable

Variable mm

åFPH&V-Ab-USD

Strip at its Bottom Level = (0.5*FPDL-H-Soil-Ab-USD + FPDL-H-Sur-Ab-USD + FPDL-H-Sur-Ab-USD)*H1*LH&V

åFPH&V-WW-USD

at its Bottom Level = (0.5*FPDL-H-Soil-WW-USD + FPDL-H-Sur-WW-USD + FPDL-H-Sur-WW-USD)*H1*LH&V

VS-V-Ab-USD

Strip = cShear-V-Ab*åFPH&V-Ab-USD

VS-H-Ab-USD

Strip on Vertical Faces= cShear-H-Ab*åFPH&V-Ab-USD

VS-V-WW-USD

= cShear-V-WW*åFPH&V-WW-USD

VS-H-WW-USD

Strip on Vertical Faces = cShear-H-WW*åFPH&V-WW-USD



Let the Clear Cover on Water Face of Abutment Wall, C-Cov.Water = 50mm, C-Cov-Water

c/de-Max.











Variable mm






Variable N-mm

Variable

0.093750

93.750/10^3

93.750*10^6

2.887

c) 270658376.983 N-mm

270.658 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)


Centroid in mm.




thus de = ds .














Snc = (b*tAb-Wall-Bot.3/12)/(tAb-Wall-Bot../2)











Abutment 5.7.3.3.2-1 M (1.33*M)Wall kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

41.797 270.658 270.658 270.658 324.790 84.104 111.858 324.790 324.790

Vertical

(-)ve Strip 69.339 270.658 270.658 270.658 324.790 130.003 172.904 324.790 324.790

Vertical

20.626 270.658 270.658 270.658 324.790 35.309 46.961 324.790 324.790

Horizontal

20.006 270.658 270.658 270.658 324.790 40.156 53.407 324.790 324.790

Horizontal

iv)


b) 0.016

2 Flexural Design of Vertical Reinforcements on Earth Face of Abutment Wall against the (-) ve Moment on Vertical Span Strip :


a) The Calculated Flexural (-) ve Moment in Vetrical Span Strip of Abutment 130.003 kN-m/m

130.003*10^6 N-mm/m

Moment for Provision of Reinforcement against (-) ve Moment value. For (-) ve 324.790 kN-m/mvalue the required Reinforcement will be on Water Face of Abutment Wall. 324.790*10^6 N-mm/m

b) 324.790 kN-m/m

324.790*10^6 N-mm/m


a) 16 mmWall.

b) 201.062


d) 27.862 mm

e) 1,347.774


for RCC Mu


(+)ve Strip

(+)ve Strip

(-)ve Strip


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


(-)åMS-V-Ab-USD


Mr

Since (-)åMS-V-Ab-USD < Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Vertical Reinforcement on Earth Face of Abutment DAb-Earth-V.

X-Sectional of 16f Bars = p*DAb-Earth-V2/4 Af-16. mm2

de-pro.

Face, dpro = (tAb-Wall-Bot.-CCov-Earth. -DAb-Earth-V./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Ab-Earth-V. mm2/m

f) 149.181 mm,C/C

g) 125 mm,C/C

h) 1,608.495


a) 0.002

b) 36.946 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 31.40 mm

c) 0.85

d) 0.047


v) Checking Against Max. Shear Force on Abutment Wall in Vetrical Strip at Bottom Level.

a) The Maximum Shear Force occurs at Botton Level of Abutment Wall on its 226.383 kN/mVertical Strip which is also the Ultimate Shearing Force for the Section. Thus 226.383*10^3 N/m

b)

b-i) 1,000.00 mm

a-ii) 600.30

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Ab-Earth-V. sreq


spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 250mm,C/C As-pro-Ab-Earth-V. mm2/m

= Af-16.b/spro



Mpro










VU.

Maximum Shear Force, VMax = VS-V-Ab-USD.= VU




the neutral axis between Resultants of the Tensile & Compressive Forces due

600.30 mm 540.000 mm

b-iii) f 0.90

b-iv) - N

c) 3,151.575 kN/m 3151.575*10^3 N/m

4,566.528 kN/m 4566.528*10^3 N/m

3,151.575 kN/m 3151.575*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied


f)its Bottom Section does not Require any Shear Reinforcement, thus Flexural Design of Vertical Reinforcement on


a) 384.923 N-mmwhere; 384.923*10^6 kN-m



h = tAb-Wall-Bot Width of Abutment Well.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-Ab-Bot of Equations as mentioned in Aritical 5.8.3.3 :










Statue between Computed Nominal Shear Resitance Vn & Factored Shearing Forces VU

For the Section (Whether Vn > VU or Vn < VU & Provisions of AASHTO-LRFD-5.8.3 have Satisfied or Not).


Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Abutment Wall on

Earth Face of Abutment Wall is OK.


427.693 N-mm 427.693*10^6 kN-m

f 0.90

b)


d) Since Abutment Wall in Vertical Direction is being considered as a Cantilever 427.693 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 427.693*10^6 N-mm

e) 130.003 kN-m

130.003*10^6 N-mm

f) Mr>Ms-v-ab-usd Satisfied


a)

Where;

b) 78.475

84.193 kN-m84.193*10^6 N-mm

1,608.495

667.00 mm Tensile Reinforcement for the Section.

c) 243.67






y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-V-Bot


Calculated Factored Moment M at Bottom of assumed Cantilever Beam (-)MS-V-Ab-USD

in Vertcal on Earth Face = (-)åMS-V-Ab-USD







i) M is Calculated Moment for the Section under Service Limit State MS-V-Ab-WSD











23,000.00 N/mm


246.000

d)

e) 7,407.341 N/mm




i)

Width Parameter, thus Provisions of Tensile Reinforcement in Vertical on Abutment Wall Earth Surface in respect

j)



a) The Calculated Flexural (+) ve Moment in Vetrical Span Strip of 84.104 kN-m/m

84.104*10^6 N-mm/m

the Governing Moment for Provision of Reinforcement against (+) ve Moment 324.790 kN-m/mvalue. For (+) ve value the required Reinforcement will be on Water Face of 324.790*10^6 N-mm/mAbutment Wall.















Since Developed Tensile Stress of Tension Reinforcement of Abutment Wall fs-Dev.< fsa Computed Tensile Stress;




Flexural Design of Vertical Reinforcements on Water Face of Abutment Wall against the (+) ve Moment on

(+)åMS-V-Ab-USD

Abutment Wall is Less than the Allowable Minimum Moment Mr. Thus Mr is

Mr

b) 324.790 kN-m/m

324.790*10^6 N-mm/m


a) 16 mmWall.

b) 201.062


d) 26.814 mm

e) 1,297.080

f) 155.011 mm,C/C

g) 125 mm,C/C

h) 1,608.495


a) 0.002

b) 36.946 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 31.404 mm

Since (+)åMS-V-Ab-USD< Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Vertical Reinforcement on Water Face of Abutment DAb-Water-V.

X-Sectional of 16f Bars = p*DAb-Water-V2/4 Af-16. mm2

de-pro.

Face, dpro = (tAb-Wall-Bot.-CCov-Water. -DAb-Water-V./2)



2))(1/2))


Spacing of Reinforcement with 16f bars = Af-16b/As-req-Ab-Water-V. sreq


spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 125mm,C/C As-pro-Ab-Water-V. mm2/m

= Af-16.b/spro



Mpro







c) 0.85

d) 0.045


iv) Since Checking have been done for Provision of Vertical Reinforcement on Earth Face of Abutment Wall against Max. Moment due to Imposed Loads & found Satisfactory in all respect thus it does not require further Checking against Provision of Vertical Reinforcement on Water Face of Abutment Wall.

4 Flexural Design of Horizontal Reinforcements on Earth Face of Abutment Wall against the (-) ve Moment on Horizontal Span Strip :


a) The Calculated Flexural (-) ve Moment in Horizontal Span Strip of 40.156 kN-m/m

40.156*10^6 N-mm/m

the Governing Moment for Provision of Reinforcement against (-) ve Moment val 324.790 kN-m/mFor (-) ve value the required Reinforcement will be on Earth Face of Abutment Wall. 324.790*10^6 N-mm/m

b) 324.790 kN-m/m

324.790*10^6 N-mm/m


a) 16 mmAbutment Wall.

b) 201.062


d) 28.577 mm

e) 1,382.400

f) 145.444 mm,C/C

g) 125 mm,C/C

h) 1,608.495





(-)åMS-H-Ab-USD


Mr

Since (-)åMS-H-Ab-USD < Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Horizontal Reinforcement on Earth Face of DAb-Earth-H.

X-Sectional of 16f Bars = p*DAb-Earth-H2/4 Af-16. mm2

de-pro.

Face, dpro = (tAb-Wall-Bot.-CCov-Earth. - DAb-Earth-V - DAb-Earth-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Ab-Earth-H. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Ab-Earth-H. sreq


spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 125mm,C/C As-pro-Ab-Earth-H. mm2/m

= Af-16..b/spro

a) 0.002

b) 36.946 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 31.404 mm

c) 0.85

d) 0.048


v) Checking Against Max. Shear Force on Abutment Wall in Horizontal Strip at Bottom Level.

a) The Maximum Shear Force occurs at Botton Level of Abutment Wall on its 39.950 kN/mHorizontal Strip which is also the Ultimate Shearing Force for the Section. Thus 39.950*10^3 N/m

b)

b-i) 1,000.000 mm


585.900 mm 540.000 mm

b-iii) f 0.90

b-iv) - N



Mpro










VU.

Maximum Shear Force, VMax = VS-H-Ab-USD.= VU






h = tAb-Wall-Bot Width of Abutment Well.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




c) 3,075.975 kN/m 3075.975*10^3 N/m

4,456.986 kN/m 4456.986*10^3 N/m

3,075.975 kN/m 3075.975*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied


f)its Bottom Section does not Require any Shear Reinforcement, thus Flexural Design of Horizontal Reinforcement on


a) 375.427 N-mmwhere; 375.427*10^6 kN-m

417.141 N-mm 417.141*10^6 kN-m

f 0.90

b)















Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Abutment Wall on

Earth Face of Abutment Wall is OK.






y(d/s-a/2)

Mn = Asfy(ds-a/2)

d) Since Abutment Wall in Horizontal Direction is being considered as Fixed End 417.141 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 417.141*10^6 N-mm

e) 40.156 kN-m

40.156*10^6 N-mm

f) Mr>Ms-h-ab-usd Satisfied


a)

Where;

b) 27.452

28.745 kN-m 28.745*10^6 N-mm

1,608.495


c) 243.667




23,000.000 N/mm

Mn-H-Bot


Calculated Factored Moment MU at Fixed Ends of assumed Rectangular (-)MS-H-Ab-USD

Beam in Horizontal on Earth Face = (-)åMS-H-Ab-USD


Factored Moment M at Bottom of Vertical Span Strip ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied).





i) M is Calculated Moment for the Section under Service Limit State (-)MS-H-Ab-WSD















246.000

d)

e) 2,591.196 N/mm




i)

Width Parameter, thus Provisions of Tensile Reinforcement in Horizontal on Abutment Wall Earth Surface in respect

j)

5on Horizontal Span Strip :


a) The Calculated Flexural (+) ve Moment in Horizontal Span Strip of 35.309 kN-m/m

35.309*10^6 N-mm/m

the Governing Moment for Provision of Reinforcement against (+) ve Moment 324.790 kN-m/mvalue. For (+) ve value the required Reinforcement will be on Water Face of 324.79*10^6 N-mm/mAbutment Wall.

b) 324.790 kN-m/m

324.79*10^6 N-mm/m


a) 16 mmofAbutment Wall.

b) 201.062





Abutment Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.





Since Developed Tensile Stress of Tension Reinforcement of Abutment Wall, fs-Dev.< fsa Computed Tensile Stress;




Flexural Design of Horizontal Reinforcements on Water Face of Abutment Wall against the (+) ve Moment

(+)åMS-H-Ab-USD


Mr

Since (+)åMS-V-Ab-USD< Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Horizontal Reinforcement on Water Face DAb-Water-H.



d) 27.475 mm

e) 1,329.065

f) 151.281 mm,C/C

g) 125 mm,C/C

h) 1,608.495

i) 0.002

j) 36.946 mm


l) Mpro>Mu OK

m) ppro<pmax OK


a) 0.45

b) c 31.40 mm

c) 0.85

d) 0.05


iv) Since Checking have been done for Provision of Horizontal Reinforcement on Abutment Wall Earth Face against Max. Moment due to Imposed Loads & found Satisfactory in all respect thus it does not require further Checking against Provision of Horizontal Reinforcement on Water Face of Abutment Wall

Part-IV :- Flexural Design of Wing Walls Against Applied Loads Under Strength Limit State (USD).

1 Computation of Related Features required for Flexural Design of Vertical & Horizontal Reinforcements on

de-pro.

Face, dpro = (tAb-Wall-Bot.-CCov-Water. -DAb-Water-V - DAb-Water-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Ab-Earth-H. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Ab-Water-H. sreq


spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 125mm,C/C As-pro-Ab-Water-H. mm2/m

= Af-16..b/spro



Mpro










Wing Walls :

i) Design Strip Width for Wing Walls in Vertical & Horizontal Direction & Clear Cover on different Faces;

a) b 1.00 m = 1000mm

b) 75 mm

50 mm



b) c Variable

c) Variable

Variable

Variable

410.00

Variable

Variable mm

Variable mm






Variable N-mm



Let the Clear Cover on Water Face of Abutment Wall, C-Cov.Water = 75mm, C-Cov-Water

c/de-Max.











Centroid in mm.




thus de = ds .









Variable

0.033750

33.750/10^3

33.750*10^6

2.887

c) 97437015.714 N-mm

97.437 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)




Wing 5.7.3.3.2-1 M (1.33*M)Walls kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

41.797 97.437 97.437 97.437 116.924 15.816 21.035 116.924 116.924

Vertical

(-)ve Strip 69.339 97.437 97.437 97.437 116.924 34.354 45.691 116.924 116.924

Vertical

9.443 97.437 97.437 97.437 116.924 16.878 22.448 116.924 116.924

Horizontal

15.572 97.437 97.437 97.437 116.924 28.374 37.737 116.924 116.924

Horizontal

iv)


b) 0.016

2 Flexural Design of Vertical Reinforcements on Earth Face of Wing Walls against the (-) ve Moment on Vertical Span Strip :







Snc = (b*tWW.3/12)/(tWW/2)










for RCC Mu


(+)ve Strip

(+)ve Strip

(-)ve Strip


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


a) The Maximum Flexural (-) ve Moment at Bottom Face of Wing Wall 34.354 kN-m/m

34.354*10^6 N-mm/m

of Moment the Reinforcement will be on Earth Face of Wing Wall. 116.924 kN-m/m 116.924*10^6 N-mm/m

b) 116.924 kN-m/m

116.924*10^6 N-mm/m


a) 16 mmWall.

b) 201.062


d) 18.305 mm

e) 885.484

f) 227.064 mm,C/C

g) 150 mm,C/C

h) 1,340.413


a) 0.004

b) 30.788 mm


d) Mpro>Mu OK

e) ppro<pmax OK


(-)åMS-V-WW-USD

is the Calcutated Factored Moment which is Less than Mr. For (-) ve value

Mr

Since (-)åMS-V-WW-USD < Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Vertical Reinforcement on Earth Face of Abutment DWW-Earth

X-Sectional of 16f Bars = p*DWW-Earth-V2/4 Af-16. mm2

de-pro.

Face, dpro = (tWW.-CCov-Earth. -DWW-Earth-V./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-WW-Earth-V. mm2/m



spro = 100mm,C/C

The provided Steel Area with 16f bars having Spacing 100mm,C/C As-pro-WW-Earth-V. mm2/m

= Af-16..b/spro



Mpro




a) 0.450

b) c 26.170 mm

c) 0.85

d) 0.071


v) Checking Against Max. Shear Force on Wing Wall in Vetrical Strip at Bottom Level.

a) The Maximum Shear Force occurs at Botton Level of Wing Wall on its 91.290 kN/mVertical Strip which is also the Ultimate Shearing Force for the Section. Thus 91.290*10^3 N/m

b)

b-i) 1,000.000 mm


330.30 mm 324.000 mm

b-iii) f 0.90

b-iv) - N

c) 1,734.075 kN/m 1734.075*10^3 N/m

2,512.617 kN/m 2512.617*10^3 N/m

1,734.075 kN/m 1734.075*10^3 N/m


c-ii) 0.000 N/m







VU.

Maximum Shear Force, VMax = VS-V-WW-USD = VU






h = tWW Thickness of Wing Wall.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-WW-Bot of Equations as mentioned in Aritical 5.8.3.3 :






c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)thus the Wing Wall does not require any Shear Reinforcement.



a) 173.909 N-mmwhere; 173.909*10^6 kN-m

193.232 N-mm 193.232*10^6 kN-m

f 0.90

b)



e) 34.354 kN-m

34.354*10^6 N-mm

f) Mr>Ms-v-ww-usd Satisfied


a)






VU For the Section (Whether Vn > VU or Vn < VU & Provisions of AASHTO-LRFD-5.8.3 have Satisfied or Not).


Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Wing Wall on its

Earth Face of Wing Wall is OK.






y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-V-Bot


Calculated Factored Moment MU at Bottom of assumed Cantilever Beam MS-V-WW-USD

in Vertcal on Earth Face = åMS-V-WW-USD


Factored Moment M at Bottom of Vertical Span Strip (Which one is Greater, if Mr ³ MU the Flexural Design for the Section has Satisfied otherwise Not Satisfied).


Where;

b) 53.065

26.104 kN-m 26.104*10^6 N-mm

1,340.413


c) 229.299




23,000.000 N/mm


246.000

d)

e) 5,322.702 N/mm





i) M is Calculated Moment for the Section under Service Limit State (-)åMS-V-WW-WSD


















Wing Wall Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.





i)

Width Parameter, thus Provisions of Tensile Reinforcement in Vertical on Wing Wall Earth Surface in respect ofControl of Cracking & Distribution of Reinforcement are OK.

j)



a) The Calculated Flexural (+) ve Moment in Vetrical Span Strip of Wing Wall 15.816 kN-m/m

15.816*10^6 N-mm/m

Moment for Provision of Reinforcement against (+) ve Moment value. For (+) 116.924 kN-m/mve value the required Reinforcement will be on Water Face of Wing Wall. 116.924*10^6 N-mm/m

b) 116.924 kN-m/m

116.924*10^6 N-mm/m


a) 16 mmWall.

b) 201.062


d) 17.082 mm

e) 826.343

f) 243.315 mm,C/C

g) 175 mm,C/C

h) 1,148.925



Since Developed Tensile Stress of Tension Reinforcement of Wing Wall fs-Dev.< fsa the Computed Tensile Stress;

the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack


Flexural Design of Vertical Reinforcements on Water Face of Wing Wall against the (+) ve Moment on

(+)åMS-V-WW

is Less than the Allowable Minimum Moment Mr. Thus Mr is the Governing

Mr

Since (+)åMS-V-WW< Mr, the Allowable Minimum Moment for the Section, thus MU

Mr is the Design Moment MU.

Let provide 16f Bars as Vertical Reinforcement on Water Face of the Wing DWW-Water-V.


de-pro.

Face, dpro = (tWW.-CCov-Water. -DWW-Water-V./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-WW-Earth-V. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-WW-Water-V. sreq


spro = 175mm,C/C

The provided Steel Area with 16f bars having Spacing 175mm,C/C As-pro-WW-Water-V. mm2/m


a) 0.003

b) 26.390 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 22.431 mm

c) 0.85

d) 0.057


iv) Since Cackings have been done for Provision of Vertical Reinforcement on Earth Face of Wing Wall against Max. Moment due to Imposed Loads & found Satisfactory in all respect, thus it does not require further Checking against Provision of Vertical Reinforcement on Water Face of Wing Wall

19 Flexural Design of Horizontal Reinforcements on Earth Face of Wing Wall against the (-) ve Moment on Horizontal Span Strip :


a) The Calculated Flexural (-) ve Moment in Horizontal Span Strip of Wing 28.374 kN-m/m

28.374*10^6 N-mm/m

Moment for Provision of Reinforcement against (-) ve Moment value. For (-) ve 116.924 kN-m/mvalue the required Reinforcement will be on Earth Face of Wing Wall. 116.924*10^6 N-mm/m

b) 116.924 kN-m/m

116.924*10^6 N-mm/m


a) 16 mm

= Af-16..b/spro



Mpro





c is the Distance between Neutral Axis & the Extrime Compressive Face,





(-)åMS-H-WW-USD


Mr

Since (-)åMS-H-WW-USD< Mr, the Allowable Minimum Moment for the Section, MU


Let provide 16f Bars as Horizontal Reinforcement on Earth Face of Wing DWW-Earth-H.

Wall.

b) 201.062


d) 19.186 mm

e) 928.126

f) 216.632 mm,C/C

g) 175 mm,C/C

h) 1,148.925


a) 0.003

b) 26.390 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 22.431 mm

c) 0.85

d) 0.064


v) Checking Against Max. Shear Force on Wing Wall in Horizontal Strip at Bottom Level.

X-Sectional of 16f Bars = p*DWW-Earth-H2/4 Af-16. mm2

de-pro.

Face, dpro = (tWW.-CCov-Earth. - DWW-Earth-V - DWW-Earth-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-WW-Earth-H. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-WW-Earth-H. sreq


spro = 175mm,C/C

The provided Steel Area with 16f bars having Spacing 175mm,C/C As-pro-WW-Earth-H. mm2/m

= Af-16..b/spro



Mpro










a) The Maximum Shear Force occurs at Botton Level of Wing Wall on its 185.346 kN/mHorizontal Strip which is also the Ultimate Shearing Force for the Section. Thus 185.346*10^3 N/m

b)

b-i) 1,000.000 mm


315.900 mm 324.000 mm

b-iii) f 0.90

b-iv) - N

c) 1,701.000 kN/m 1701.000*10^3 N/m

2,464.693 kN/m 2464.693*10^3 N/m

1,701.000 kN/m 1701.000*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)

VU.

Maximum Shear Force, VMax = VS-H-WW.= VU






h = tWW Thickness of Wing Wall.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h

















thus the Abutment Wall does not require any Shear Reinforcement.

f)its Earth Face does not Require any Shear Reinforcement, thus Flexural Design of Horizontal Reinforcement on


a) 143.214 N-mmwhere; 143.214*10^6 kN-m

159.126 N-mm 159.126*10^6 kN-m

f 0.90

b)


d) Since Wing Wall in Horizontal Direction is being considered as Fixed End 159.126 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 159.126*10^6 N-mm

e) 28.374 kN-m

28.374*10^6 N-mm

f) Mr>Mu Satisfied


a)

Where;

b) 45.721

18.438 kN-m 18.438*10^6 N-mm

1,148.925

351.000 mm

Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Wing Wall on

Earth Face of Wing Wall is OK.






y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-H-Bot


Calculated Factored Moment MU at Fixed Ends of assumed Rectangular åMS-H-WW-USD

Beam in Horizontal on Earth Face = åMS-H-WW-USD


Factored Moment MU on Horiziontal Strip Span ( Which one is Greater, if Mr ³ MU the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit åMS-H-WW-WSD



the Tensile Reinforcement for the Section.

c) 217.814




23,000.000 N/mm


246.000

d)

e) 4,827.918 N/mm




i)

j)






Wing Wall, CCov-Earth = 75mm & Bar Dia, DBar = 16f ; thus dc = (16/2 + 50)mm















Since Developed Tensile Stress of Tension Reinforcement of Well Cap fs-Dev.< fsa the Computed Tensile Stress;

the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Horizontal on Wing Wall Earth Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.


15on Horizontal Span Strip :


a) The Calculated Flexural (+) ve Moment in Horizontal Span Strip of Wing 16.878 kN-m/m

16.878*10^6 N-mm/m

Moment for Provision of Reinforcement against (+) ve Moment value. For (+) ve 116.924 kN-m/mvalue the required Reinforcement will be on Water Face of Wing Wall. 116.924*10^6 N-mm/m

b) 116.924 kN-m/m

116.924*10^6 N-mm/m


a) 16 mmWall.

b) 201.062


d) 17.845 mm

e) 863.219

f) 232.921 mm,C/C

g) 175 mm,C/C

h) 1,148.925

i) 0.003

j) 26.390 mm


l) Mpro>Mu OK

m) ppro<pmax Not OK

Flexural Design of Horizontal Reinforcements on Water Face of Wing Wall against the (+) ve Moment

(+)åMS-H-WW


Mr

Since (+)åMS-V-WW< Mr, the Allowable Minimum Moment for the Section, thus MU


Let provide 16f Bars as Horizontal Reinforcement on Water Face of Wing DWW-Water-H.

X-Sectional of 16f Bars = p*DWW-Water-H2/4 Af-16. mm2

de-pro.

Face, dpro = (tWW.-CCov-Water. -DWW-Water-V - DWW-Water-H./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-WW-Earth-H. mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Ab-Water-H. sreq


spro = 175mm,C/C

The provided Steel Area with 16f bars having Spacing 175mm,C/C As-pro-Ab-Water-H. mm2/m

= Af-16..b/spro



Mpro





a) 0.450

b) c 22.43 mm

c) 0.85

d) 0.060


iv) Since Checking have been done for Provision of Horizontal Reinforcement on Earth Face of Wing Wall against Max. Moment due to Imposed Loads & found Satisfactory in all respect thus it does not required further Checking against Provision of Horizontal Reinforcement on Water Face of Wing Wall







) & Live

Maximum

Maximum

Mr)

M.


2 Structural Type, Design Criteria & Dimentions of Structure :



iii) Design Criteria : Service Limit State (WSD) Design According to AASHTO-LRFD-2004.

iv) Dimensions of Substructural Components :


a) Height of Wing Wall from Bottom of Well Cap up to its Top, H 6.147 m

Structural Design of Wing Wall Counterfort :

C

C

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

522512

0020

00

6350

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

RL-5.00m

RL-2.20m

3000 2100

b) Height of Wing Wall from Top of Well Cap up to its Top, H1 4.947 m







n) Thickness of Counterfort Wall For Wing Wall. 0.450 m


p) Length of Counterfort Base (Length at Bottom). 3.450 m

q) Clear Spacing between Counerfort & Abutment Wall at Bottom 1.775 m

r) Average Spacing between Counerfort & Abutment Wall 2.375 m

s) 2.825 m

t) Thickness of Wing Walls within Well Cap, 0.450 m

u) Length of Cantilever Wing Walls (On Top). 3.000 m

v) Thickness of Cantilever Wing Walls 0.450 m

w) Height of Rectangular Portion of Cantilever Wing Walls 2.000 m

x) Height of Triangular Portion of Cantilever Wing Walls 1.500 m

y) Longitudinal Length of Well Cap on Toe Side from Abutment Wall 2.525 mOuter Face.

3 Design Data in Respect of Unit Weight, Strength of Materials & USD Multiplier Factors :

i)

9.807


hWell-Cap.

hSteam.

H-W-Wall

LW-W-Well-Cap

t.-Ab-wall-Bot.

t.-Ab-wall-Top.

tWW-Countf.

NW-W-count

LCount-Base.


SAver-Count&Ab.



tWW

LCant-WW.

tCant-WW

hCant.-WW-Rec.

hCant.-WW-Tri.

LW-Cap-Toe.



gc kg/m3





ii)







a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.89

e) 2.89 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa


a) 8.384 n 8

b) r 20c) k 0.29d) j 0.90

e) R 1.10





gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


c



c Ec
















d) Multiplier Factor for Vertical Earth Pressure on Substructure Components of 1.000 Bridge-EV; Applicable to Abutment & Wing Walls, (Max. value of Table 3.4.1-2)





b) 1.000

c) IM 1.000 ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1; SERVICE - I(Applicable only for Truck Loading & Tandem Loading)

d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000



gEH


gEV

gES



Multiplier Factor for Vehicular Dynamic Load Allowence-IM as per Provision of















q) -

r) -

t) 1.000



a) 0.441

b) f 34



d) 0.34 to 0.45 dim



0.007935


c) 0.018

18.000














O


Dp-ES kN/m2






kN/m2

0.004761

4.761



0.004761

4.761

c) k 0.441

d) 1,835.424

e) g 9.807

f) 900.000 mm




h) 1,050.000 mm



600.000 mm 0.600 m

6 Philosophy in Flexural Design of Counterforts for Bridge substructure :

a) On Extrme edge of Well Cap's Earth Side Counterforts are being proposed for Bridge Substructure as an additional

Dimensions with Effective Span Length of Counterfort & its Base Length. The Components will also have to Face the Horizontal Dead Load (DL ) & Live Load (LL) Pressures caused by same Forces acting upon Vertica Faces of Wing Walls having Dimensions with Effective Span Length of Counterforts & Height of Wing Wall.

b) In normal cases, Counterforts would have Well Cap/Abutment Cap & Wing Wall/Abutment Wall on their both sides, but for the present case Wall Cap is existing beyond Counterfort. At the same time beyond Well Cap there exists

Vertical Span would be considered from Counterfort Outer Face up to the Middle Point between Abutment Wall and

Length from Middle Point between Abutment Wall and Counterfort + Equivelent Horizontal Length of Cantilever Wing

Dp-LS = kgEgheq*10-9 Dp-LL-Ab³6.00m N/mm2

kN/m2


kN/m2

Dp-LS = kgEgheq*10-9 , Dp-LL-WW³6.00m N/mm2

kN/m2






Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.

Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

Support for the Cantilever Wing Wall Components. These Counterforts will have to Face Vertical Dead Loads (DL) due to Soil & Surcharge Pressure including Live Load (LL) Surcharge Pressure upon an Area of Well Cap having the

a Cantilever Type Wing Wall at Top Level of Counterfort. Thus for Calculation of Vertical Loads (DL & LL) Effective

the Counterfort. Whereas for Calculation of Horizontal Loads (DL & LL) the Effective Horizontal Span would be the

Wall.

c) Since the Counterfort posses Triangular shape & the Applied Forces are of Complex Nature, thus it is considered to

d) The Main Reinforcements of Counterfort will be Against Applied Horizontal Forces upon Effective Horizontal Span of Counterfort. The Shear/Web Reinforcements would be under both Vertical & Horizontal Forces accordingly.

d) To reduce the Complex in Calculation of Horizontal Loads upon the 1.668 mTrapezoid Shape Cantilever Wing Wall, keeping its Height & Vertical Surface Area same, an Equivalent Horizontal Length is being considered for the Component. Thus the Calculated Equivalent Horizontal Length of Cantilever

e) For Flexural Design & Calculation of Loads, Shear & Moments due to Applied 1.000 m

1.000m Width/Depth Strips are being considered both in Vertical & Horizontal. 1.000 m

7

i) Skech Diagram Showing Dimentions Related to Loads Upon Counterfort :

0.45 m

A

q

6.147 m 4.947 m

B FC G

1.2 mD E

2.525 m

3.45 m

ii) Computation of Span Lengths for Vertical & Horizontal Loads upon Counterfort :

adopt the Service Limit State Methodology (WSD) in Flexural Design of Counterforts for Wing Wall.

LCant-WW-Equi.

Wing Wall = (LCant-WW.*hCant.-WW-Rec. + 0.50*LCant-WW.*hCant.-WW-Tri.)/H1

bStrip-V

hStrip-H

Calculation of Loads upon Counterfort:

tWW. =

H = H1 =

90o K/

K

G/

q

hWell-Cap. =

LW-Cap-Toe.. =

LCount-Base. =

G/

a) Clear span in between Counterfort & Abutment Wall (Average), 2.375 m

b) Thickness of Counterfort for Wing Wall, 0.450 m

c) Effective Span of Counterfort for Horizontal Load Calculations; 3.305 m

d) Effective Span of Counterfort for Vertical Load Calculations; 1.638 m

iii) Computation of Horizontal Loads acting upon Counterfort :

a) Intensity of Horizontal Load acting upon Counterfort on its Bottom Level due to 7.208 kN/mActive Horizontal Earth Dead Load Pressure upon Horizontal Effective Span

b) Intensity of Horizontal Load acting upon Counterfort on its Bottom Level due to 26.225 kN/mthe Horizontal Surcharge Dead Load Pressure upon Horizontal Effective Span

c) Intensity of Horizontal Load acting upon Counterfort on its Bottom Level due to 27.535 kN/mthe Horizontal Surcharge Live Load Pressure for Wing Wall upon Horizontal

d) Intensity of Total Horizontal Loads acting upon Counterfort on its Bottom Level 60.968 kN/m

iv) Computation of Factored Horizontal Loads acting upon Counterfort :

a) Intensity of Factored Horizontal Load acting upon Counterfort on its Bottom 7.208 kN/mLevel due to Active Horizontal Earth Dead Load Pressure upon Horizontal

b) Intensity of Factored Horizontal Load acting upon Counterfort on its Bottom 26.225 kN/mLevel due to the Horizontal Surcharge Dead Load Pressure upon Horizontal

c) Intensity of Horizontal Load acting upon Counterfort at its Bottom due to the 60.968 kN/mHorizontal Surcharge Live Load Pressure for Wing Wall upon Horizontal

d) Intensity of Total Factored Horizontal Loads acting upon Counterfort on its 94.400 kN/m

SAver-Count&Ab.

tWW-Countf.

SEff.-Coutf-H-Load.

= SAver-Count&Ab./2+ tWW-Countf.+ LCant-WW-Equi.

SEff.-Coutf-V-Load.

= SAver-Count&Ab./2+ tWW-Countf.

pEarth-DL-H

Length; = ko*H1*SEff.-Coutf-H-Load.hStrip-H

pSur-DL-H

Length; = Dp-ES*SEff.-Coutf-H-Load.hStrip-H

pSur-LL-H

Effective Span Length; = Dp-LL-WW*SEff.-Coutf-H-Load.hStrip-H (H1<6.000m).

pH-(DL+LL)

due to all Horizontal Loads/Pressure (DL & LL) upon the Horizontal Effective

Span Length; = pEarth-DL-H + pSur-DL-H + pSur-LL-H

FpEarth-DL-H

Effective Span Length; = gDE*pEarth-DL-H

FpSur-DL-H

Effective Span Length; = gDS*pSur-DL-H

FpSur-LL-H

Effective Span Length; = gLL-LS.*pSur-LL-H . (H1<6.000m).

FpH-(DL+LL)

Bottom Level due to all Horizontal Loads/Pressure (DL & LL) upon the

Horizontal Effective Span Length; = FpEarth-DL-H + FpSur-DL-H + FpSur-LL-H

v) Computation of Vertical Loads acting upon Counterfort :

a) 145.813 due to Active Vertical Earth Dead Load Pressure upon Vertical Effective Span

b) 47.160 kN/mdue to Dead Load Pressure caused by Self Weight of Well Cap upon Vertical

c) 13.642 kN/mdue to Vertical Surcharge Live Load Pressure for Wing Wall upon Vertical

d) Intensity of Total Horizontal Loads acting upon Counterfort on its Bottom Level 206.615 kN/m

vi) Computation of Factored Vertical Loads acting upon Counterfort :

a) 145.813

b) 47.160 kN/m

c) 13.642 kN/m

d) Intensity of Total Factored Vertical Loads acting upon Counterfort on its 206.615 kN/m

8

i) Selection of Plane for Critical Section to Calculate action of Shearing Forces & Moments :

a)

b) Since the Counterfort is of Triangular Shape having Max. Horizontal Forces on its Bottom Level, thus to Calculate the Effective Depth of Flexural Reinforcement on Tension Face, the Perpendicular Distance from Earth Face of theCounterfort up to the Critical Point should be determined.

c) GF 3.450 m

Intensity of Vertical Load acting upon Counterfort Bottom Level at F & G pEarth-DL-V kN/m

Length; = wE*H1*SEff.-Coutf-V-Load.hStrip-V

Intensity of Vertical Load acting upon Counterfort Bottom Level at F & G pSelf.-Wt.-DL-V

Effective Span Length; = wC*hWell-Cap*SEff.-Coutf-V-Load.hStrip-V.

Intensity of Vertical Load acting upon Counterfort Bottom Level at F & G pSur-LL-V

Effective Span Length; = Dp-LL-WW*SEff.-Coutf-V-Load.hStrip-V. (H1<6.000m).

pH-(DL+LL)

due to all Horizontal Loads/Pressure (DL & LL) upon the Horizontal Effective

Span Length; = pEarth-DL-V+ pSelf.-Wt.-DL-V + pSur-LL-V

Intensity of Factored Vertical Load acting upon Counterfort Bottom Level at FpEarth-DL-V kN/m F &r G due to Active Vertical Earth Dead Load Pressure upon Vertical Effective

Span Length; = gDV*pEarth-DL-V

Intensity of Factored Vertical Load acting upon Counterfort Bottom Level at FpSelf.-Wt.-DL-V

F & G due to Dead Load Pressure caused by Self Weight of Well Cap upon

Vertical Effective Span Length; = gDC*pSelf.-Wt.-DL-V

Intensity of Factored Vertical Load acting upon Counterfort Bottom Level at FpSur-LL-V

F & G due to Vertical Surcharge Live Load Pressure for Wing Wall upon

Vertical Effective Span Length; = gLL-LS.*pSur-LL-V. (H1<6000m).

FpV-(DL+LL)

Bottom Level at F & G due to all Vertical Loads/Pressure (DL & LL) upon the

Horizontal Effective Span Length; = FpEarth-DL-V + FpSelf.-Wt.-DL-V + FpSur-LL-V

Calculation of Shearing & Moments Forces acting upon Counterfort due to Applied Forces :

The Critical Section of Counterfort Prevails at B, the Point on Toe Face of Wing Wall at Top Surface of Well Cap.

From Sketch Diagram Distance between Point G & F = LCount-Base.

d) BF 3.900 m

e) Height of Wing Wall from Top of Well Cap up to its Top, H1 4.947 m

g)

h) 0.697

i) q 34.892 0 0.609 rad

j) 0.572

k) 0.820

l) GK 2.830 m

m)

n) Acoordingly from Geometry of Sketch Diagram Design Depth of Counterfort 3.199 m

ii) Calculation of Shearing Forces acting upon Critical Section of Counterfort due to Applied Forces :

a) 614.561 kN

b) 712.821 kN

iii)

a) 58.796 kN-m/m

b) 320.900 kN-m/m

c) 746.025 kN-m/m

From Sketch Diagram Distance between Point B & F = LCount-Base. + tWW

Let qo be the Angle formed at Top between the Earth Side Vertical Face of Wing Wall & Counterfort's Inclined Face.

From Sketch Diagramvalue of tanq = GF/H1 tanq

From the Calcutated value of tanq, the value of q = tan-1

Thus Value of sinq = sin34.4880 sinq

Value of cosq = cos34.4880 cosq

From Geometry of Sketch Diagram the Perpendicular Distance from Point G up to Inclined Face of Counterfort, GK = AG sinq = H1*sinq

Let Consider G/ a Point on Wing Wall Vertical Face on a Line BK/ which is Parallel to GK & Drown up to Inclined

Inclined Face of Counterfort. Since the Parallel Lines GK & BK/ are very colse, thus it may Consider GK = G/K/

BK/

on its Bottom Level is BK/ = GK + tWW*cosq .

The Horizontal Shearing Force on Counterfort Critical Section at B is the Total VCritical-H

Factored Horizontal Loads (DL & LL) on its Bottom Level for the Effective

Horizontal Span Length = 0.50*FpEarth-DL-H*H1 + FpSur.-DL-H*H1 + FpSur-LL-H*H1

The Vertical Shearing Force on Counterfort Critical Section at B is the Total VCritical-V

Factored Vertical Loads (DL & LL) on its Bottom Level for the Effective Verticla

Span Length = FpV-(DL-LL)*LCount-Base.

Calculation of Design Moments for Provision of Reinforcements Against Applied Horizontal Forces:

Bending Moment at B due to Factored Horizontal Dead Load Earth Pressure MEarth-DL-H

= 1/3*FpEarth-DL-H *H12

Bending Moment at B due to Factored Horizontal Dead Load Surcharge MSur-DL-H

Pressure = 0.50*FpSur-DL-H *H12

Bending Moment at B due to Factored Horizontal Live Load Surcharge MSur-LL-H

Pressure = 0.50*FpSur-LL-H*H12

d) 1,125.721 kN-m/m

1125.721*10^6 N-mm/m

9 Flexural Design of Reinforcements for Counterfort at Bottom Level against Calculated Factored Moment :

i) Provisions of Different Elements for Design & Checking for Effective Depth of Counterfort on its Bottom:

a) b 450 mm

b) Since the all Faces of Counterfort will be Covered by Earth, thus 75mm Clear 75 mmCovers are being proposed on its all Faces.

c) 25 mmthe Calculated Moments at Bottom Level of Counterforts

d) 490.874

d) Let the Spacing between 2-Layers of Main Reinforcement Bars = 32mm 32 mm

e) With 2-Layers of Main Bars, Clear Cover & Bar Layer Spacing the Provided 3,082.918 mmEffective Depth for Tensial Reinforcements of Counterfort on its Bottom Leval

f) 1,506.338 mm

g)

ii) Design of main Reinforcement for Counterfort against Calculated Factored Moment on its Bottom Level:

a) 2,465.371

b) 5.022 nos.

c) 8 nos.Main Reinforcements for the Counterforts in 2-Layers havin 4nos. Bars on EachLayer.

d) 3,926.991

iii) Checking of Design for Reinforcement in respect of Shear:

a) V 405.687 kNFace of Counterfort & Water Face of Wing Wall), thus Formula for the Effective 405.687*10^3 N

Total Factored Bending Moment at B due to all applied Forces MTotal-H

= MEarth-DL-H + MSur-DL-H + MSur-LL-H

Width of Counterfort for Flexural Design, b = 450mm

CCover.

Let Consider - Layers 25f Bars will be required as Flexural Reinforcement for DBar-Main.

X-Sectional Area of 25f Main Bars = p*DBar2/4 Af-25 mm2

sBar-Layers.

dpro-B.

at B; = BG/ - CCover - sBer-Layers/2 - DBar

Since the Counterfort acts as a T-Beam, thus against the Calculated Factored dreq-B.

Moment on Bottom Level of Counterfort at B the Required Effective Depth;

= (MTotal-H /Rb)1/2

Since the Provided Effective Depth for Tensial Reinforcement dpro > dreq, the Calculated Effective Depth, thus the Ptovision of Counterfort Depth on its Bottom Level is OK.

Required Steel Area against Calculated Moment = MTotal-H/jdfs As-req mm2

No's of 25f bars required = As-req/Af-25 NBars-req.

As per Convention & Practice of Design, Let provide with 8 No's 25f bars as NBars-pro.

Provided Steel Area againest Provided Main Rinforcements = Af-25*NBars-pro As-pro-Main mm2

Since the Counterfort T-Beam posses variable Depth (Distance between Earth

b) 3,758.577 mm

c) 0.240

d) p 0.232

e) 0.222

f)Provide Reinforcements as well as Shrinkage & Temperature Reinforcements for Un-Reinforced Surfaces.

v) Curtailment of Main Reinforcements in Counterforts:

a) Since Horizontal Earth Pressure on Counterforts reduces with reduction of Depth in upward direction and Bending

b) 3.604 kN/m

c) 26.225 kN/m

d) 23.602 kN/m

e) 53.430 kN/m

f) 308.524 kN-m/m

g)

Wing Walls have a 3.000 m long Trapezium Shape Extended Cantilever Part at Top Level, whose Self Weight and Imposed Loads should be supported by the Counterfort. Thus it is Recommended not reduce the Number of Main Bars of Counterfort.

10

Shearing Force on Bottom Level V = VTotal-H - (MTotal-H/d/)*tanq; where,

d/ is Horizontal Effective Depth of Counterfort having value = dpro/cosq d/

The Shearing Stress due to calculated Shearing Force V & Effective Depth d/ tv N/mm2

= V/bd/

Percentage of Steel Provided for the section = 100*As/bd/ N/mm2

With p = 0.147, f/C =21MPa, the Permissible Shearing Stress in concrete, tc N/mm2

Computed from Table 5.1 of Book Treasure of RCC Design (In S.I Units) ofSushik Kumar.

Since tv > tc thus Shear Reinforcements in the form of Vertical and Horizontal Ties are required for Counterforts to

Moment also reduces accordingly. For these reduction of Moments in respect of Depth (h), the requrment of Steel

Area for the Section also reduces. No of Bars required for a Section µ h2, the Depth of Section - h.

At Depth 1/2H1 from Top the Intensity of Factored Horizontal Earth Dead Fp1/2-Earth-DL-H

Load Pressure; = gDEko*1/2H1*SEff.-Coutf-H-Load.*hStrip-H

At Depth 1/2H1 from Top the Intensity of Factored Horizontal Surcharge Fp1/2-Sur-DL-H

Dead Load Pressure; = gDSDp-ES*SEff.-Coutf-H-Load.*hStrip-H

At Depth 1/2H1 from Top the Intensity of Factored Horizontal Surcharge Live Fp1/2-Sur-LL-H

Load Pressure; = gLL-LS.Dp-LL-WW*SEff.-Coutf-H-Load.*hStrip-H

Intensity of Total Factored Horizontal Loads acting upon Counterfort at Fp1/2H-(DL+LL)

1/2H1 from Top due to Horizontal Loads (DL & LL) upon the Horizontal Effective

Span Length; = Fp1/2-Earth-DL-H + Fpp1/2-Sur-DL-H+ Fpp1/2-Sur-LL-H

Factored Miment due to Horizonal Loads at a Depth 1/2H1 for Section MH-1/2H

= 0.50*Fp1/2-Earth-DL-H*1/3*(1/2H1)2 + Fp1/2-Sur-DL-H*(1/2*H1)2 -Fp1/2-Sur-LL-H*(1/2*H1)2

From Calculations Sl.-8-v-(f) it Appaires that the Calculated Moment at a Depth 1/2H1 is about 1/3rd of Moments at Bottom Level. On the mentioned ground 50% of Main Reiforcement from Depth 1/2H1 Depth can be Cut off. But the

Design of Shear-Web (Tie) Reiforcements for Counterforts:

i)

a) 94.400 kN/m 94.400*10^3 N/m

b) 53.430 kN/m 53.430*10^3 N/m

c) 10 mm

d) 157.080

e) 575.611

f) 272.892 mm c/c

g) 200 mm c/c

ii) Design of Vertical Ties for Counterforts:

a) Sketck Diagram Showing Vertical Load Upon Counterfort :

0.45 m

A

q

6.147 m 4.947 m

B FC G

1.2 mD E

Design of Horizontal Ties for Counterforts:

Factored Horizontal Shearing Force at Depth H1 from Top due to Horizontal VH1-(DL+LL)

Loads/Pressure (DL & LL) acting upon Counterfort for its Horizontal Effective

Span Length; = FpH-(DL+ LL)

Factored Horizontal Shearing Force at Depth 1/2H1 from Top due to V1/2H1-(DL+LL)

Horizontal Loads/Pressure (DL & LL) acting upon Counterfort for its Horizontal

Effective Span Length; = FpH-(DL+ LL)

Let Provide 2-Legged 10f bars as Horizontal Ties for the Counterfort, DBar-Hor.

X-Sectional of 2-Legged 10f bars as Horizontal Ties = 2*pDBar-Hor2/4 Af-10 mm2

Steel Area required for Horizontal Ties against Horizontal Forces VH1-(DL+LL) As-req-Tie-H1 mm2

at Depth H1 from Top of Wing Wall = VH1-(DL+LL)/fs

Spacing required for 2-Legged Horizontal Ties at Depth H1 from Top of Wing sreq.-Tie-H1

Wall = Af-10*hStrip-H/As.req-H1.

Let Provide 200mm C/C spacing for 2-Legged Horizontal Ties with 10f spro-Tie-H1-1/2H1

bars for Total Depth.

tWW. =

H = H1 =

90o K/

K

G/

q

hWell-Cap. =

2.525 m

3.45 m

206.6147 kN/m 206.6147 kN/m

b) 206.615 kN/m

N/m

c) 206.615 kN/m

N/m

d) Since the Vertically Factored Shearing Forces at Points F & G have same Magnitude, thus it requires to ProvideShearing/Web Reinforcements for its full Horizontal Length.

e) 12 mm

f) 226.195

e) 1,259.846

e) 179.542 mm

f) 150 mm C/C

LW-Cap-Toe.. =

LCount-Base. =

FpV-(DL+LL)-G = FpV-(DL+LL) -F =

Vertically Downward Factored Shearing Force at Point F due to Earth & Self VD-F

Weight of Well Cap Dead Loads (DL)and Suecharge Live Load (LL) = FpV-(DL+LL) 206.615*10^3

Vertically Downward Factored Shearing Force at Point G due to Earth & Self VD-G

Weight of Well Cap Dead Loads (DL)and Suecharge Live Load (LL) = FpV-(DL+LL) 206.615*10^3

Let Provide 2-Legged 12f bars as Horizontal Ties for the Counterfort, DBar-Hor.

X-Sectional of 2-Legged 12f bars as Horizontal Ties = 2*pDBar-Hor2/4 Af-12 mm2

Area of Steel required as Vertical Ties for Counterforts at Points F & G As-req-V-Tie mm2

= VD-F&G/fs

Spacing for 2-Legged Vertical Ties with 12f in between F & G sreq-Vert.F-G

= Af-12*hStrip-V/As-req-V-Tie

Let Provide 150mm C/C Spacing for Vertical Ties with 2-Legged 12f bars spro-Virt.(F& G)

in between Points F & G of Counterfort.

Dimensions with Effective Span Length of Counterfort & its Base Length. The Components will also have to Face the

Length from Middle Point between Abutment Wall and Counterfort + Equivelent Horizontal Length of Cantilever Wing

) Surcharge Pressure upon an Area of Well Cap having the

DESIGN OF SUB-STRUCTURE FOR CHAMPATOLI BRIDGE AT 11.90km ON BAGHAIHUT-MACHALONG-SEZEK ROAD

N.


2 Structural Type, Design Criteria & Dimensions of Structure :



iii) Design Criteria : Ultimate Stress Design (AASHTO-LRFD-2004).

iv) Dimensions of Substructure Components :



Structural Design of Cantilever Portion of Wing Walls:

C

C

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

5225

1200

2000

6350

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

RL-5.00m

RL-2.20m

3000 2100








h) Inner Distance in between Wing Walls (Transverse), 9.350 m

g) Thickness of Wing Wall Counterforts 0.450 m

h) Number of Wing-Wall Counterforts (on each side) 1.000 No's

i) Thickness of Wing Walls within Well Cap, 0.450 m

j) Thickness of Cantilever Wing Walls 0.450 m

k) Horizontal Length of Cantilever Wing Walls 3.000 m

l) Height of Rectangular Portion of Cantilever Wing Walls 2.000 m

m) Height of Triangular Portion of Cantilever Wing Walls 1.500 m

3 Design Data in Respect of Unit Weight, Strength of Materials & Load Multiplier Factors :

i)

9.807





e) Unit weight of Earth (Compacted Clay/Sand/Silt) 1,835.424

ii)




hWell-Cap.

hSteam.

hBack-wall

H-W-Wall

LW-W-Well-Cap

LAb-T-Inner

tWW-Countf.

NW-W-count

tWW

tCant-WW

LCanti-WW

hCanti-WW-Rec.

hCanti-WW-Trian.


(Having value of Gravitational Acceleration, g m/sec2)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3



e) Unit weight of Earth (Compacted Clay/Sand/Silt) 18.000


a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

g) 200000.000 MPa


a) 8.384 n 8

b) r 20c) k 0.29 d) j 0.90

e) R 1.10





d) For Axil Compression with Spirals or Ties & Seismic Zones at Extreme 0.750 Limit State (Zone 3 & 4).


f) For Compression in Strut-and-Tie Models 0.700




j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85

4 Different Load Multiplying Factors for Strength Limit State Design (USD) & Load Combination :

i) Permanent & Dead Load Multiplier Factors under Strength Limit State Design (USD):

wW-Sali. kN/m3

wE kN/m3


c



c Ec







Steel Allowable Strength under Service Load Condition (SLC) = 0.40fy fs






fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.











ii) Live Load Multiplier Factors :a) Multiplier Factor for Multiple Presence of Live Load ( No of Lane = 2)-m m 1.000

(ASSHTO LRFD-3.6.1.1.1)

b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000



gEH


gEV


gES



















q) -

r) -

t) 1.000







d) Multiplier Factor for Vertical Earth Pressure on Substructure Components of 1.000 Bridge-EV; Applicable to Abutment & Wing Walls, (Max. value of Table 3.4.1-2)



ii) Live Load Multiplier Factors for Service Limit State Design (WSD) According to AASHTO-LRFD-3.4.1;











gEH


gEV

gES



Table 3.4.1-1&2 :


b) 1.000


d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






q) -


Multiplier Factor for Vehicular Dynamic Load Allowence-IM as per Provision of

















r) -

t) 1.000



a) 0.441

b) f 34.000

c) Angle of Friction with Concrete surface & Soil d 19 to 24


d) 0.34 to 0.45 dim


a) Constant Horizontal Earth Pressure due to Uniform Surcharge, 7.935

0.008


c) 0.018

18.000


a) Constant Earth Pressure both Vertical & Horizontal for Live Load 0.007


0.005

4.761

b) Constant Horizontal Earth Pressure due to Live Load Surcharge for 0.008331


0.004761

4.761

c) k 0.441






Effective Friction Angle of Soil, f = 340 .(Table 12.9, Page-138, RAINA's Book)

O


Dp-ES kN/m2


ks is Coefficient of Earth Pressure due to Surcharge = ko for Active ks

qs is Uniform Surcharge applied to upper surface of Active Earth Wedge (Mpa) wE*10-3 N/mm2



kN/m2


kN/m2


kN/m2


kN/m2

ks is Coefficient of Lateral Earth Pressure = ko for Active Earth Pressure.


d) 1,835.424

e) g 9.807

f) 900.000 mm




h) 1,050.000 mm



600.000 mm 0.600 m

7 Philosophy in Flexural Design of Cantilever Wing Walls of Bridge Structure :

a)

Structure having Fixed End on Bottom in Horizontal Plane. But the Cantilever Wing Walls have Vertical Fixed Enddue to their Trapezoidal Shape having Variable Depth. These Wing Walls also have to Face the same Horizontal

Walls those should Designed as a Cantilever Flexural Component against the affect of Horizontal Loads & Forces

b) Cantilever Wing Walls have Vertical Fixed End & their Depth gradually reduces with distance from Support. Load

those gradually reduce with reduction of Depth. But the Cumulative Horizontal Loads on Upper Strip is Higher thanthat of Deeper Strip due to greater Span Length. Since in Cantilever Component Max. Moment occurs on SupportEdge due to Total Applied Loads, thus the Flexural Design should be done considering the Cantilever Wing Walls are of Span Strips in Horizontal Direction Each having 1.000m Depth. On Rectangular Portion Loads will the Totalfor Span & on Triangular Portion those will the Average of Top & Bottom Span Strips.

c) 1.000 m

Forces & Moments Let considered the Cantilever Wing Walls being divided in 0.500 m4 (Four) Horizontal Span Strips of whom Top 3 (Three) have 1.000m Depth &the Bottom One is of 0.500m Depth.

d) Sketch Diagram of Cantilever Wing Wall & Arrangement of Design Strips :





Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.

Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

The Wing Walls of Bridge Substructure will have to face Horizontal Dead Load (DL) & Live Load (LL) Pressures due to Earth & Surcharge upon them. Under Horizontal Pressure (DL & LL) the Wing Walls behave as Cantilever

Loads/Pressure (DL & LL) as that of Normal Wing Walls. Thus to ensure the sustainability of these Type of Wing

(DL & LL) accordingly.

Intensity caused by Horizontal Earth Pressures (DL) & Surcharge Loads (DL & LL) are Higher at Deep Depth &

For the Design purpose & Calculation of Imposed Loads (DL & LL), Shearing DV-Strip-1-3

DV-Strip-4


3 m

1.00 m

3 m 2.000 m

1.00 m

3 m

1.00 m

2 m 1.500 m

0.50 m

0.500 m

1.000 m

8 Calculations of Horizontal Loads Upon Design Strips of Cantilever Wing Wall:

i) Design Strips & Strip Dimensions :

a) Let the Design Strips in Horizontal Direction with having Depth in Vertical 1.000 m

0.500 m

b) Total Number of Horizontal Strips for Design in Cantilever Wing Walls 4.000 nos.

c) Horizontal Length of 1st. Strips on Wing Wall's Rectangular Portion 3.000 m

d) Horizontal Length of 2nd Strips on Wing Wall's Rectangular Portion 3.000 m

e) Average Horizontal Length of 3rd Strip on Wing Wall's Triangular Portion 2.000 m(Considering Equivalent Rectangular Shape for the Strip).

f) Average Horizontal Length of 4th Strip on Wing Wall's Triangular Portion 0.500 m(Considering Equivalent Rectangular Shape for the Strip).

ii) Intensity of Horizontal Earth & Surcharge Pressure (DL & LL) on Wing Wall Components :

a) 7.935

Constant Factor for Site Soil Condition.

b) 7.935

a Constant Factor for Site Soil Condition.

LCanti-WW =

DV-Strip-1 =

LSteip-1 = hCanti-WW-Rec. =

DV-Strip-2=

LSteip-2 =

DV-Strip-3 =

LSteip-(Aver)-3 = hCanti-WW-Trian. =

DV-Strip-4 =

LSteip-(Aver)-4 =

LSteip-Top-4 =

DV-Strip-1-3.

Direction D = 1.000m for Top 3-Strips & for Bottom One D = 0.500m DV-Strip-4

NStrip.

L-Strip-1

L-Strip-2

L-Strip-(Aver)-3.

L-Strip-(Aver)-4.

Intensity of Horizontal Earth Pressure at any Depth h, on Cantilever Wing wE*ko kN/m3

Wall can Expressed by pEarth = wE*ko*h ; where wEko in kN/m3 ; Which is a

Intensity of Surcharge Dead Load Horizontal Pressure at any Depth h, on Dp-ES kN/m2

Cantilever Wing Wall can Expressed by pSur-DL = Dp-ES in kN/m2 ; Which is


c) 8.331

d) Sketch Diagram showing Depth of Different Levels Horizontal Strips from Top of Cantilever Wing Wall.

3 m

1.00 m

1.000 m

3 m

1.00 m 2.000 m

3 m

1.00 m 3.000 m

2 m

0.50 m 3.500 m

0.500 m

1.000 m

iii) Total Horizontal Dead Load (DL) Earth Pressure on Different Strips of Cantilever Wing Wall :(Considering Intensity of Horizontal Earth Pressure on Bottom of Strip is Effective on full Depth).

a) 23.804 kN

b) 47.607 kN

c) 47.607 kN

d) 6.943 kN

iv) Total Horizontal Dead Load (DL) Surcharge Pressure on Different Strips of Cantilever Wing Wall :

a) 23.804 kN

b) 23.804 kN

Intensity of Surcharge Live Load Horizontal Pressure at any Depth h, on Dp-LL-WW kN/m2

Cantilever Wing Wall can Expressed by pSur-LL-WW = Dp-LL-WW in kN/m2 ; Which is a Constant Factor for Site Soil Condition & Depth h £ 6.000m ³ .For Present Case Depth h < 6.000m.

LCanti-WW =

DV-Strip-1 =

hStrip-1-Bot. =

LSteip-1 =

DV-Strip-2= hStrip-2-Bot. =

LSteip-2 =

DV-Strip-3 = hStrip-3-Bot. =

LSteip-(Aver)-3 =

DV-Strip-4 = hStrip-4-Bot. =

LSteip-(Aver)-4 =

LSteip-Top-4 =

Total Horizontal Earth Pressure on Strip-1 PE-Strip-1

= wEko*hStrip-1-Bot*DV-Strip-1*LStrip-1


= wEko*hStrip-2-Bot*DV-Strip-2*LStrip-2


= wEko*hStrip-3-Bot*DV-Strip-3*LStrip-(Aver)-3


= wEko*hStrip-4-Bot*DV-Strip-4*LStrip-(Aver)-4

Total Horizontal Dead Load (DL) Surcharge Pressure on Strip-1 PSur-DL-Strip-1

= Dp-ES*DV-Strip-1*LStrip-1



c) 15.869 kN

d) 1.984 kN

v) Total Horizontal Live Load (LL) Surcharge Pressure on Different Strips of Cantilever Wing Wall :

a) 24.993 kN

b) 24.993 kN

c) 16.662 kN

d) 2.083 kN

vi) Total Horizontal Dead Loads (DL) due to Earth & Surcharge Pressure on Different Strips :

a) 47.607 kN

b) 71.411 kN

c) 63.476 kN

d) 8.926 kN

9 Calculations of Factored Horizontal Loads Upon Design Strips of Cantilever Wing Wall under Strength Limit State Design (USD):

i) Factored (USD) Horizontal Dead Load (DL) Earth Pressure on Different Strips :

a) 35.705 kN

b) 71.411 kN

= Dp-ES*DV-Strip-2*LStrip-2


= Dp-ES*DV-Strip-3*LStrip-(Aver)-3


= Dp-ES*DV-Strip-4*LStrip-(Aver)-4

Total Horizontal Live Load (LL) Surcharge Pressure on Strip-1 PSur-LL-Strip-1

= Dp-LL-WW*DV-Strip-1*LStrip-1


= Dp-LL-WW*DV-Strip-2*LStrip-2


= Dp-LL-WW*DV-Strip-3*LStrip-(Aver)-3

Total Horizontal Live Load (DL) Surcharge Pressure on Strip-4 PSur-LL-Strip-4

= Dp-LL-WW*DV-Strip-4*LStrip-(Aver)-4

Total Horizontal Dead Load due to Earth & Surcharge Pressure on Strip-1 PDL-Strip-1

= PE-Strip-1 + PSur-DL-Strip-1


= PE-Strip-2 + PSur-DL-Strip-2


= PE-Strip-3+ PSur-DL-Strip-3


= PE-Strip-4+ PSur-DL-Strip-4

Factored Horizontal Earth Pressure on Strip-1 = gEH*PE-Strip-1 PE-Strip-1-USD

Factored Horizontal Earth Pressure on Strip-2= gEH*PE-Strip-2 PE-Strip-2-USD


c) 71.411 kN

d) 10.414 kN

ii) Factored (USD) Horizontal Dead Load (DL) Surcharge Pressure on Different Strips :

a) 35.71 kN

b) 35.705 kN

c) 23.804 kN

d) 2.975 kN

iii) Factored (USD) Horizontal Live Load (LL) Surcharge Pressure on Different Strips :

a) 43.737 kN

b) 43.737 kN

c) 29.158 kN

d) 3.645 kN

iv) Factored (USD) Total Horizontal Loads (DL & LL) on Different Strips of Cantilever Wing Wall :

a) 115.148 kN

b) 150.853 kN

c) 124.373 kN

d) 17.034 kN



Factored Horizontal Dead Load (DL) Surcharge Pressure on Strip-1 PSur-DL-Strip-1-USD

= gES*PSur-DL-Strip-1

Total Horizontal Dead Load (DL) Surcharge Pressure on Strip-2 PSur-DL-Strip-2-USD



= gES*PSurDL-Strip-3



Factored Horizontal Live Load (LL) Surcharge Pressure on Strip-1 PSur-LL-Strip-1-USD

= gLL-LS*PSur-LL-Strip-1




=gLL-LS*PSur-LL-Strip-3

Factored Horizontal Live Load (DL) Surcharge Pressure on Strip-4 PSur-LL-Strip-4-USD


Factored Total Horizontal Loads (DL & LL) on Strip-1 åPStrip-1-USD

= PE-Strip-1-USD + PSur-DL-Strip-1-USD + PSur-LL-Strip-1-USD








10 Calculations of Factored Horizontal Loads Upon Design Strips of Cantilever Wing Wall under ServiceLimit State Design (WSD):

i) Factored (WSD) Horizontal Dead Load (DL) Earth Pressure on Different Strips :

a) 23.804 kN

b) 47.607 kN

c) 47.607 kN

d) 6.943 kN

ii) Factored (WSD) Horizontal Dead Load (DL) Surcharge Pressure on Different Strips :

a) 23.804 kN

b) 23.804 kN

c) 15.869 kN

d) 1.984 kN

iii) Factored (WSD) Horizontal Live Load (LL) Surcharge Pressure on Different Strips :

a) 24.993 kN

b) 24.993 kN

c) 16.662 kN

d) 2.083 kN

iv) Factored (WSD) Total Horizontal Loads (DL & LL) on Different Strips of Cantilever Wing Wall :

a) 72.600 kN

Factored Horizontal Earth Pressure on Strip-1 = gEH*PE-Strip-1 PE-Strip-1-WSD

Factored Horizontal Earth Pressure on Strip-2= gEH*PE-Strip-2 PE-Strip-2-WSD



Factored Horizontal Dead Load (DL) Surcharge Pressure on Strip-1 PSur-DL-Strip-1-WSD


Total Horizontal Dead Load (DL) Surcharge Pressure on Strip-2 PSur-DL-Strip-2-WSD



= gES*PSurDL-Strip-3



Factored Horizontal Live Load (LL) Surcharge Pressure on Strip-1 PSur-LL-Strip-1-WSD





=gLL-LS*PSur-LL-Strip-3

Factored Horizontal Live Load (DL) Surcharge Pressure on Strip-4 PSur-LL-Strip-4-WSD


Factored Total Horizontal Loads (DL & LL) on Strip-1 åPStrip-1-WSD


b) 96.403 kN

c) 80.138 kN

d) 11.009 kN

11 Calculation of Moments on Different Strips of Cantilever Wing Wall at Counterfort Face ;

i) Moments on Different Strips at Counterfort Face due Factored Loads under Strength Limit Stale (USD) :

a) 172.722 kN-m

172.722*10^6 N-mm

b) 226.280 kN-m

226.280*10^6 N-mm

c) 124.373 kN-m

124.373*10^6 N-mm

d) 4.259 kN-m

4.259*10^6 N-mm

ii) Moments on Different Strips at Counterfort Face due Factored Loads under Service Limit Stale (WSD) :

a) 108.900 kN-m

108.900*10^6 N-mm

b) 144.605 kN-m

144.605*10^6 N-mm

c) 80.138 kN-m

80.138*10^6 N-mm

d) 2.752 kN-m

2.752*10^6 N-mm

iii) Moments on Different Strips at Counterfort Face due Unfactored Dead Loads :

a) 71.411 kN-m

71.411*10^6 N-mm

= PE-Strip-1-WSD + PSur-DL-Strip-1-WSD + PSur-LL-Strip-1-WSD







Moment at Counterfort Face due to Factored Loads on Strip-1 MStrip-1-USD

= åPStrip-1-USD*LStrip-1/2


= åPStrip-2-USD*LStrip-2/2


= åPStrip-3-USD*LStrip-(Aver)-3/2


= åPStrip-4-USD*LStrip-(Aver)-4/2

Moment at Counterfort Face due to Factored Loads on Strip-1 MStrip-1-WSD

= åPStrip-1-WSD*LStrip-1/2


= åPStrip-2-WSD*LStrip-2/2


= åPStrip-3-WSD*LStrip-(Aver)-3/2


= åPStrip-4-WSD*LStrip-(Aver)-4/2

Moment at Counterfort Face due to Factored Loads on Strip-1 MDL-Strip-1-UF

= PDL-Strip-1*LStrip-1/2


b) 107.116 kN-m

107116*10^6 N-mm

c) 63.476 kN-m

63.476*10^6 N-mm

d) 2.232 kN-m

2.232*10^6 N-mm

12 Computation of Related Features required for Flexural Design of Main Reinforcements (Horizontal) for

i) Design Strip Width for Cantilever Wing Walls in Vertical Direction & Clear Cover on different Faces;

a) b-1 1.00 mb-2 1.00 mb-3 1.00 mb-4 0.50 m

b) 75 mm

50 mm



b) c Variable

c) Variable

Variable

Variable

-

Variable

Variable mm

Variable mm



= PDL-Strip-2*LStrip-2/2


= PDL-Srip-3*LStrip-(Aver)-3/2


= PDL-Strip-4*LStrip-(Aver)-4/2

Cantilever Wing Walls :

Let Consider the Design Width in Vertical direction for Strip-1 = 1000mmStrip-2 = 1000mmStrip-3 = 1000mmStrip-4 = 500mm


Let the Clear Cover on Water Face, C-Cov.Water = 50mm, C-Cov-Water

c/de-Max.











Centroid in mm.




thus de = ds .






Variable N-mm

Variable

0.033750

33.750/10^3

33.750*10^6

2.887

c) 97437015.714 N-mm

97.437 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g) 0.017

16.875/10^3

16.875*10^6

h) 48.719 kN-m48718507.857 N-mm

g)














Snc = (b*tCant-WW.3/12)/(tCant-WW./2). (For Strip-1, Strip-2 & Strip-3).


For Nonprestressing & Monolithic or Noncomposite Beam or Elements, Mcr-Strip-1,2,3

Sc = Snc & fcpe = 0, thus Equation for Cracking Moment Stands to Mcr = Sncfr (For Strip-1, Strip-2 & Strip-3).




For Strip-4 having Strip Width b = 0.500m; Snc m3

Snc = (b*tCant-WW.3/12)/(tCant-WW./2). m3

mm3

For Strip-4 having Strip Width b = 0.500m; Mcr = Sncfr Mcr-Strip-4

Table-1 Showing Allowable Resistance Moment M r for Minimum Reinforcement of Strips Direction


Strip No. Value of Value of Actuat Acceptable M Maximum

for Unfactored Cracking Factored Allowable FlexuralMoments Dead Load As per Moment Cracking Cracking Moment Factored Min. Moment Moment

on Cant. Moment Equation Value Moment Moment of Section Moment

Wing 5.7.3.3.2-1 M (1.33*M)Wall kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

Strip-1 71.411 97.437 97.437 97.437 116.924 172.722 229.720 116.924 172.722

Strip-2 107.116 97.437 97.437 97.437 116.924 226.280 300.953 116.924 226.280

Strip-3 63.476 97.437 97.437 97.437 116.924 124.373 165.415 116.924 124.373

2.232 48.719 48.719 48.719 58.462 4.259 5.664 58.462 58.462

iv)


b) 0.016

13 Flexural Design of Horizontal Reinforcements on Earth Face of Cantilever Wing Wall against Calculated the Moments on Strips-1 :


a) 172.722 kN-m/m

172.722*10^6 N-mm/m

Governing Moment for Provision of Reinforcement against Moment value. 116.924 kN-m/mSince the Strip is facing Earth Loads, thus requird Reinforcements will be on 116.924*10^6 N-mm/mEarth Side.

b) 172.722 kN-m/m

172.722*10^6 N-mm/m


a) 16 mm

b) 201.062


d) 27.388 mm



for RCC Mu


Strip-4


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


Calculated Flexural Moment in Horizontal Span Strip-1 of Cantilever Wing MStrip-1-USD

Wall is Greater than the Allowable Minimum Moment Mr. Thus MStrip-1 is the

Mr

Since MStrip-1-USD> Mr, the Allowable Minimum Moment for the Section, thus MU

MStrip-1 is the Design Moment MU.

Let provide 16f Bars as Horizontal Reinforcement on Earth Face of Strip-1. DStrip-1

X-Sectional of 16f Bars = p*DStrip-12/4 Af-16. mm2

de-pro.

Face, dpro = (tCant.-WW -CCov-Earth. -DStrip-1/2)



2))(1/2))


e) 1,324.861

f) 151.761 mm,C/C

g) 100 mm,C/C

h) 2,010.619


a) 0.005

b) 46.182 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 39.255 mm

c) 0.85

d) 0.107


v) Checking Against Calculated Shear Force at Counterfot Face of Cantilever Wing Wall on Strip-1 :

a) 115.148 kN/mTotal Factored Load on the Strip which is also the Ultimate Shearing Force for 115.148*10^3 N/m

b)

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Strip-1-Earth mm2/m

Spacing of Reinforcement with 16f bars = Af-16b/As-req-Strip-1-Earth sreq


spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 125mm,C/C As-pro-Strip-1-Earth mm2/m

= Af-16.b/spro



Mpro










The Shear Force at Counterfot Face of Cantilever Wing Wall on Strip-1 is the VU.

the Section. Thus Shear Force, VStrip-1 = åPStrip-1-USD = VU



b-i) 1,000.000 mm


330.300 mm 324.000 mm

b-iii) f 0.90

b-iv) - N

c) 1,734.075 kN/m 1734.075*10^3 N/m

2,512.617 kN/m 2152.617*10^3 N/m

1,734.075 kN/m 1734.075*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)

f)





h = tCant-WW Thickness of Cant. Wing Wall.Thus value 0.9*de for the Section; 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




The Nominal Shear Resitance Vn for the Section is the Lesser value of any Vn-Strip-1 of Equations as mentioned in Aritical 5.8.3.3 :












Since Nominal Shear Resitance for the Section Vn > VU the Calculated Ultimate Shearing Force for the Section,thus the Cantilever Wing Wall on Strip-1 does not require any Shear Reinforcement.

Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Cantilever Wing Wall on Strip-1 Section does not Require any Shear Reinforcement, thus Flexural Design for Provisio of Horizontal Reinforcement on Earth Face of Cantilever Wing Wall on Strip-1 is OK.



a) 255.152 N-mmwhere; 256.152*10^6 kN-m

283.503 N-mm 283.503*10^6 kN-m

f 0.90

b)


d) 283.503 kN-mBeam having 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 283.503*10^6 N-mm

e) 172.722 kN-m

172.722*10^6 N-mm

f) Mr>MStrip-1-USD Satisfied


a)

Where;

b) 147.581

108.900 kN-m 108.900*10^6 N-mm

2,010.619


c) 262.482






y(d/s-a/2)

Mn = Asfy(ds-a/2)

Since Cantilever Wing Wall on Strip-1 is being considered as a Cantilever Mn-Strip-1

at its Support Face will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment MU at Support Face of Cantilever Beam is on MStrip-1-USD

its Earth Face = MStrip-1-USD


Factored Moment M at Support Face ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit MStrip-1-WSD








23,000.000 N/mm


246.000

d)

e) 12,931.833 N/mm




i)

j)















the Cant. Wing Wall Structure, thus value of Crack Width Parameter Z should calculate based the value of fs-Dve.





Since Developed Tensile Stress of Tension Reinforcement of Cant. Wing Wall fs-Dev.< fsa Computed Tensile Stress;

though Computed Tensile Stress fsa>0.6fy; but Developed Crack Width Parameter ZDev.<ZMax. Allowable Max.Crack Width Parameter, thus Provisions of Tensile Reinforcement in Cant. Wing Wall Strip-1 Earth Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.





a) 226.280 kN-m/m

226.280*10^6 N-mm/m

Governing Moment for Provision of Reinforcement against Moment value. 116.924 kN-m/mSince the Strip is facing Earth Loads, thus requird Reinforcements will be on . 116.924*10^6 N-mm/mEarth Side

b) 226.280 kN-m/m

226.280*10^6 N-mm/m


a) 16 mm

b) 201.062


d) 36.341 mm

e) 1,757.951

f) 114.373 mm,C/C

g) 75 mm,C/C

h) 2,680.826


a) 0.007

b) 61.576 mm




Mr

Since MStrip-2-USD > Mr, the Allowable Minimum Moment for the Section, thus MU

MStrip-2-USD is the Design Moment MU.



de-pro.




2))(1/2))


Spacing of Reinforcement with 16f bars = Af-16b/As-req-Strip-2. sreq


spro = 100mm,C/C


= Af-16.b/spro



Mpro



d) Mpro>Mu OK

e) pmax>ppro OK


a) 0.450

b) c 52.340 mm

c) 0.85

d) 0.143




b)

b-i) 1,000.000 mm


330.300 mm 324.000 mm

b-iii) f 0.900

b-iv) - N

c) 1,734.075 kN/m 1734.075*10^3 N/m

2,512.617 kN/m










the Section. Thus Shear Force, VStrip-2 = PStrip-2-USD = VU













2152.617*10^3 N/m

1,734.075 kN/m 1734.075*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.00

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)

f)


a) 332.589 N-mmwhere; 255.152*10^6 kN-m

369.543 N-mm 283.503*10^6 kN-m

f 0.90

b)




















y(d/s-a/2)

Mn = Asfy(ds-a/2)



e) 226.280 kN-m

226.280*10^6 N-mm



a)

Where;

b) 146.98

144.605 kN-m 144.605*10^6 N-mm

2,680.826


c) 288.899








Factored Moment M at Support Face ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State MStrip-2-WSD














23,000.000 N/mm


246.000

d)

e) 11,701.238 N/mm




i)

j)



a) 124.373 kN-m/m

124.373*10^6 N-mm/m


b) 124.373 kN-m/m

124.373*10^6 N-mm/m


a) 16.000 mm












though Computed Tensile Stress fsa>0.6fy;but Developed Crack Width Parameter ZDev.<ZMax. Allowable Max.Crack Width Parameter, thus Provisions of Tensile Reinforcement in Cant. Wing Wall Strip-2 Earth Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.




Mr

Since MStrip-3-USD> Mr, the Allowable Minimum Moment for the Section, thus MU

MStrip-3 is the Design Moment MU.



b) 201.062


d) 19.504 mm

e) 943.470

f) 213.109 mm,C/C

g) 175 mm,C/C

h) 1,148.925


a) 0.003

b) 26.390 mm


d) Mpro>Mu OK

e) pmax>ppro OK


a) 0.450

b) c 22.431 mm

c) 0.85

d) 0.061



de-pro.




2))(1/2))


Spacing of Reinforcement with 16f bars = Af-16b/As-req-Strip-3 sreq


spro = 175mm,C/C


= Af-16.b/spro



Mpro













b)

b-i) 1,000.000 mm


330.300 mm 324.000 mm

b-iii) f 0.90

b-iv) - N

c) 1,734.075 kN/m1734.075*10^ N/m

2,512.617 kN/m 2152.617*10^3 N/m

1,734.075 kN/m1734.075*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N


the Section. Thus Shear Force, VStrip-3= PStrip-3-USD = VU





















d) Vn>Vu Satisfied

e)

f)


a) 149.997 N-mmwhere; 173.909*10^6 kN-m

166.663 N-mm193.232*10^6 kN-m

f 0.90

b)



e) 124.373 kN-m

124.373*10^6 N-mm



a)

Where;

b) 190.056










y(d/s-a/2)

Mn = Asfy(ds-a/2)






Factored Moment MStrip-2-USD at Support Face ( Which one is Greater, if Mr ³ MStrip-2-USD

the Flexural Design for the Section has Satisfied otherwise Not Satisfied)






80.138 kN-m80.138*10^6 N-mm

1,148.925


c) 217.814




23,000.000 N/mm


246

d)

e) 20,068.839 N/mm



i) M is Calculated Moment for the Section under Service Limit State of Loads MStrip-3-WSD
























i)

j)



a) 4.259 kN-m/m

4.259*10^6 N-mm/m


b) 58.462 kN-m/m

58.462*10^6 N-mm/m


a) 16 mm

b) 201.062


d) 18.305 mm

e) 442.742

f) 227.064 mm,C/C

g) 175 mm,C/C

h) 574.463



the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allowable Max.Crack Width Parameter, thus Provisions of Tensile Reinforcement in Cant. Wing Wall Strip-3 Earth Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.



Wall is Less than the Allowable Minimum Moment Mr. Thus Mr is the

Mr

Since MStrip-4 -USD < Mr, the Allowable Minimum Moment for the Section,thus MU




de-pro.




2))(1/2))


Spacing of Reinforcement with 16f bars = Af-16b/As-req-Strip-4 sreq


spro = 175mm,C/C




a) 0.003

b) 26.390 mm


d) Mpro>Mu OK

e) pmax>ppro OK


a) 0.450

b) c 22.431 mm

c) 0.85

d) 0.061




b)

b-i) 500.000 mm


330.300 mm 324.000 mm

= Af-16.b/spro



Mpro











the Section. Thus Shear Force, VStrip-4 = åFPStrip-4 -USD = VU








b-iii) f 0.90

b-iv) - N

c) 867.038 kN/m867.038*10^3 N/m

1,256.309 kN/m 1256.309*10^3 N/m

867.038 kN/m867.038*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)

f)


a) 74.998 N-mmwhere; 74.998*10^6 kN-m

83.332 N-mm83.332*10^6 kN-m





















f 0.90

b)



e) 4.259 kN-m

4.259*10^6 N-mm



a)

Where;

b) 13.055

2.752 kN-m 2.752*10^6 N-mm

574.463


c) 217.814





y(d/s-a/2)

Mn = Asfy(ds-a/2)






Factored Moment MStrip-4-USD at Support Face ( Which one is Greater, if Mr ³ MStrip-4-USD

the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State of Loads MStrip-4-WSD












23,000.000 N/mm


246

d)

e) 1,378.492 N/mm




i)

17 Provision of Shrinkage & Temperature Reinforcements both on Earth & Water Faces of Cantilever Wing

i) Requiment of Shrinkage & Temperature Reinforcements for Cantilever Wing Walls :

a) The Flexural Design of Cantilever Wing Walls are being done Considering the Structure as Constituents of Multiple Cantilever Beam Strips having Span Length in Longitudinal, Width in Vertical & Depth in Transverse Direction. Sincethe Cantilever Wing Walls are Subject to Horizontal Earth & Surcharge Pressures, thus under Concept of Flexural

all Concrete Surfaces should have Reinforcements in both Directions either under Flexural Tension/Compression or under Flexural Shear/Web/Torsion. The Surfaces where there is no such type of Reinforcements on that surface
















the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allowable Max.Crack Width Parameter, thus Provisions of Tensile Reinforcement in Cant. Wing Wall Strip-4 Earth Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.

Walls in Vertical & Horizontal Directions :

Design the Longitudinal Reinforcements are on Earth Face in Vertical Plane. According to Principle of RCC Design


b) On Earth Surface Cantilever Wing Walls the Main Reinforcements are in Horizontal Direction, thus on Earth Face Shrinkage & Temperature Reinforcements are required in Vertical Direction. Whereas on Water Face Provision of Shrinkage & Temperature Reinforcements are required both in Vertical & Horizontal Directions.

ii) Provision of Shrinkage & Temperature Reinforcements on Earth Face in Vertical and for Wate Face both in Vertical & Horizontal Direction :

a) Since the Thickness/Depth of Cantilever Wing Walls is less than1200mm, 1.000 m

thus to Calculate the Shrinkage & Temperature Reinforcements on both 1.000 mFaces in Vertical & Horizontal Directions a Strip is being Considered having b 1.000 m

c) 268.293 Temperature Reinforcement for Structural Components having its Thickness

d) 1000000.000

e) 16 mmEarth Face Both in Vertical & Horizontal Direction on Water Face.

f) 201.062

g) 749.413 nos.Direction on Earth Face Both Vertical & Horizontal Direction on Water

h)Reinforcements should not Spaced further Apart than 3.00 Times the Component's Thickness or 450mm.

1,350.000 mm

ii) Allowable Max. Spacing for Shrinkage & Temperature Reinforcements 450.000 mm

i) 300 mm

Direction on Water Face.

j) 670.206

k)1200mm Thickness, the Spacing of Shrinkage & Temperature Reinforcements Bars should not Exceed 300mm inEach Direction on all Faces and Steel Area of Shrinkage & Temperature Reinforcements need not Exceed value of

Shrinkage & Temperature Reinforcements should arrange according to AASHTO-LRFD-5.10.8.1. Provisions.

LHor.

hVer.

Length of each Arm b = 1000mm.

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T mm2

1200mm or Less; As ³ 0.11Ag/fy in both way.(Here Thickness = 450mm).

Here Ag is Gross Area of Strip on Vertical Surface = LHor.*hVer. Ag-Strip mm2

Let provide 16f bars as Shrinkage & Temperature in Vertical Direction on DBar-S&T-V&H

X-Sectional Area of 16f bar = pDBar-S&T-V&H2/4 Af-16 mm2

Spacings required for 16f Bars as Shrinkage & Temperature in Vertical sreq-S&T-V&H

Face. = Af-16*b/As-req-S&T-V&H

According to AASHTO-LRFD-5.10.8.1. In a Component having Less 1200mm Thickness, Shrinkage & Temperature

i) 3.00 Times of Cantilever Wing Wall Thickness = 3.00*tCont.-WW 3.00*tCont.-WW

sAllow-S&T-V&H-1

Let provide 300 mm Spacing for Shrinkage & Temperature Reinforcements spro-S&T-V&H

with 16f Bars in Vertical Direction on Earth Face, Both in Vertical & Horizontal

The provided Steel Area with 16f Bars as Shrinkage & Temperature As-pro-S&T-V&H mm2/m

Reinforcements having Spacing 300mm,C/C = Af-16.b/spro-S&T-V&H

According to AASHTO-LRFD-5.10.8.1. For Components of Solid Structural Concrete Wall & Footing having Less

åAb = 0.0015Ag. Since the Cantilever Wing Walls are Concrete Wall Structure, thus


i) Allowable Max. Spacing for Shrinkage & Temperature Reinforcements 300.000 mm

1,500.000

i) Status between Provided Steel Area of Shrinkage & Temperature Reinforcement & Allowable Max, Steel Area for

Provisions of Shrinkage & Temperature Reinforcement have Satisfied, otherwise Not Satisfied)åAb > As-pro-S&T Satisfied

l)Shrinkage & Temperature on Surfaces of Cantilever Wing Wall is OK.

sAllow-S&T-V&H-2

ii) Calculated value of åAb = 0.0015Ag. åAb = åAb = mm2/m

Shrinkage & Temperature Reinforcement (Whether åAb > As-pro-S&T-V&H or not. If åAb < As-pro-S&T-V&H ; then

Since Calculated åAb > As-pro-S&T-V&H. > As-req-S&T-V&H. & spro-S&T-V&H. = sAllow-S&T-V&H-2, thus Provisions for the


Not Satisfy

O. Structural Design of Abutment Well Cap :

1

2 Dimension of Different Sub-Structural Components & RCC Well for Foundation:


i)







C

C

Dimensions of Sub-Structure.

hWell-Cap.

hSteam.

hb-wall

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

5225

1200

2000

6350

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

10250

9350

12750

3450 3450 3450600 600 600600

34503450

60 0

2750

60 0

2525

1775

5500

3000

450

450

450

450

2150

2150

2450

2750

RL-5.00m

RL-2.20m

3000 2100

300


g) Width of Wall Cap on Heel Side (From Abutment Wall Face). 2.975 m

h) Width (Longitudinal Length) of Abutment Well Cap, 5.500 m

i) Length (Transverse Length) of Abutment Well Cap, 12.750 m

j) 10.250 mDirection.

k) Inner Length of Abutment Wall in between Wing Walls (Transverse), 9.350 m







r) 2.825 m








z) Surface Area of Well Cap 66.879

H-W-Wall

WWell-Cap-Hell

LAb-Cap-Long.

LW-Cap-Trans

Transverse Length of Abutment Wall (Outer Face to Outer Face) in X-X LAB-Wall-Trans.

LAb-T-Inner

t.-Ab-wal-Bot.

t.-Ab-wal-Top.

tWW-Countf.

NW-W-count


SAver-Count&Ab.



t-Wing-wall

tw-wall-Cant.

Lw-wall-Cant.

hw-wall-Cant.-Rec.

hw-wall-Cant.-Tri.

L-W-Cap-Toe.

L-W-Cap-Heel-Aver.


AWell-Cap. m2

ii) Dimensions of RCC Well for Foundation.

a) 5.500 m

b) 12.750 m





h) 4.300 m

i) 7.250 m

j) 4.300 m





o) 4.900 m

p) 4.050 m

3

a) Type of Sub-soil : a) At Borehole No-BH07 (Cox's Bazar End), from GL (GL is - 2.25m from Road Top Level) up to 2.50m depth Sub-soil posses Loss gray fine Silty sand having SPT Value ranging 7 to 12. Whereas in next 0.75m from depth 2.50m to 3.75m there exists Medium dense gray fine sand with SPT value ranging from 12 to 40. From depth about 3.75m there exists Bed-rock (Gray Shale) having 50 and over SPT values .b) At Borehole No-BH08 (Teknuf End), from GL (GL is - 2.25m from Road Top Level) upto 2.75m depth Sub-soil posses Medium dency gray sandy silt having



HWell-pro.

tWell.

tWall-Perti

DOuter.




NPock.




Effective Span Length in Y-Y Direction (C/C Distance between Well Walls.) SEff-Y-Y.

= SPock-Y-Y. + tWall

Effective Span Length in X-X Direction (C/C Distance between Well Walls.) SEff-X-X

= SPock-X-X-Central/Outer + tWall(tWall-Perti)

Information about Soil, Foundation, Abutment & Wing-walls:

SPT Value renging 12 to 37. In next 2.15m (About depth 2.75m to 4.80m) there exists Medum densey gray fine sand with SPT value renging from 37 to 50.From depth about 4.80m there exists Bed rock (Gray Shale) having SPT value 50 over.

b) Type of Foundation : Due to its Geographical position, Marin Drive Road have every risk to effected by Wave action & Cyclonic Strom from Sea. More over the Slain Water is also an important factor for RCC Construction Works in these area. In Designing of any Permanent Bridge/Structure on this Road, specially in Foundation Design all the prevalling adverse situations should be considered for their Survival andDurability. Though as per Soil Investigation Report there exist Loss to Mediumdency gray sandy silt on Seashore Sub-soil, but due to ground their formation those posses a very poor Mechanical bonding among it contitutent.But there exites Bed-rock at a considerably short depth (About 3.75m to 4.80m) from the Ground Level.Presence of Bed-rock is an important for the Foundation of any Structue on this Road. To encounter all mentioned adverse situations Provisionof RCC Caissons embedded into the Bed-rock will be best one as Foundation of Bridges on this Road. RCC Caissons embedded into the Bed-rock will be aSolid mass to save guard the Structure against Errosion, Sliding, Overturning etc. which caused by the Wave action & Cyclonic Strom. More over against Salinity effect necessary meassary can provide for RCC Caissions. Thus it isrecommended to Provide RCC Caissions embedded into the Bed-rock at least1.50m into Bed-rock as Foundation of Delpara Bridge.

c) Type of Abutment : Wall Type Abutment.

d) Type of Wing-walls : Wall Type Wing Walls Integrated with Abutment Wall having Counterforts over Well & Cantilever Wings beyond Well.

e) Design Criteria : Strength Limit State of Design (USD) According to AASHTO-LRFD-2004.

4 Design Data in Respect of Unit Weight, Flexural Multiplier Factors, Material Strength & Soil Pressure:

i)

(Having value of Gravitional Acceleration, g = 9.807






ii)

a) Unit weight of Normal Concrete 24.000 kN/m^3

b) Unit weight of Wearing Course 23.000 kN/m^3

c) Unit weight of Normal Water 10.000 kN/m^3


m/sec2)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc

wWC

wW-Nor.

d) Unit weight of Saline Water 10.250 kN/m^3

e) Unit weight of Earth (Compected Clay/Sand/Silt) 18.000 kN/m^3

iii) Design Data for Resistance Factors for Conventional Construction (AASHTO LRFD-5.5.4.2.1). :










j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85

vi) Strength Data related to Ultimate Strength Design( USD & AASHTO-LRFD-2004) :

a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa

v) Strength Data related to Working Stress Design & Service Load Condition ( WSD & AASHTO-SLS ) :

a) 8.384 n 8

b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102

vi) Sub-soil Investigation Report & Side Codition Data:

a) N 50 Over

b) 33

wW-Sali.

wE


fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.




c



c Ec










Value of Ratio of Steel & Concrete Flexural Strength, r = fs/fc Value of k = n/(n + r) = 9.000/(9.000 + 20) Value of j = 1 - k/3 = 1 - 0.307/3


kN/m3



c) Recommended Allowable Bearing Capacity of Soil as per Soil Investigation p 770

5 Intensity of Different Imposed Loads, Load Coefficients & Multiplier Factors :


a) 0.441

b) f 34

c) Angle of Friction with Concrete surface & Soli d 19 to 24AASHTO-LRFD-3.11.5.3 ;Table 3.11.5.3-1.

d) 0.34 to 0.45 dim



0.007935


c) 0.018




0.004761 4.761

b) Constant Horizontal Earth Pressur due to Live Load Surcharge for 0.00833 Wing Walls (Parallel to Traffic), Where; 8.331

0.004761 4.761

c) k 0.441

d) 1,835.424

e) g 9.807

= 15 + 1/2(50 - 15) = 32.5 . Say N/ = 33

kN/m2

Report witht SPT Value 50 over, p = 7.2 Ton/ft2. = 770kN/m2





O


Dp-ES kN/m2




= wE*10-3N/mm2


kN/m2


kN/m2


kN/m2


kN/m2


gsis Unit Weight of Soil (kg/m3) gs kg/m3


f) 900.000 mm 600.000 mm




i) 1,050.000 mm 600.000 mm


j) Width of Live Load Surcharge Pressure for Wing Walls, 600.000 mm 0.600 m 600.000 mm 0.600 m


a) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.0008 MpaElements in Lateral Direction of Wind Flow (Parallel to Traffic). 0.800 AASHTO-LRFD-3.8.1.2.2; Table-3.8.1.2.2-1.

b) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.0009 MpaElements in Longitudinal Direction of Wind Flow (Perpendicular to Traffic). 0.900


a) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.000950 Mpa 0.950

b) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.001645 MpaElements in Longitudinal Direction (Perpendicular to Traffic). 1.645







Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm. Weq-WW³6.00m.

pWind-Sup-Let.

kN/m2

pWind-Sup-Long.

kN/m2

pWind-Sub-Let.



pWind-Sub-Long.

kN/m2


pWind-LL-Sup-Let.


pWind-LL-Sup-Long.

action at 1800mm above Deck & Considering 600 Skew Angle of Main




i) Permanent & Dead Load Multiplier Factors under Limit State (USD) according to AASHTO-LRFD-3.4.1:








ii) Live Load Multiplier Factors under Limit State (USD) according to AASHTO-LRFD-3.4.1:


b) 1.750

c) IM 1.330

(Applicable for Truck Loading & Tandem Loading only)

d) 1.750

e) 1.750

f) 1.750

Force; for Two Lane Bridge.(AASHTO-LRFD-3.8.1.3; Table- 3.8.1.3-1).

pLL-Sup-Break.

pLL-25%-Truck.

pLL-5%-(Tru+Lane.)



gEH


gEV


gES



Multiplier Factor for Vhecular Dynamic Load Allowence-IM as per Provision of ASSHTO LRFD-3.6.2.1, Table 3.6.2.1-1;




g) 1.750

h) 1.750

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

7 Different Load Multiplying Factors for Service Limit State Design (USD) & Load Combination :






















gEH







b) 1.000


d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000





gEV


gES










Multiplier Factor for Wind Load on Structure-WS gLL-WS

Multiplier Factor for Wind Load on Live Load-WL gLL-W

Multiplier Factor for Water Load & Stream Pressure-FR gLL-FR






q) -

r) -

t) 1.000

8 Calculations of Vertical Loads on Abutment Well Cap, Moments & Shearing Forces under Limite State Condition (USD) :

i) Philosophy in Computation of different Loads (DL & LL) upon Well Cap:

a) Since the Well Cap will have to receive all Vertical Loads both from Live Load Elements & Dead Loads from the Superstructure & Substructure Components, Soil Load, Surcharge Load and its Self Weight, thus computation of these Loads should done accordingly.

b)(Perpendicular to Traffic), thus all Loads from Superstructure & Loads of Abutment Wall including Components

these Uniformly Distributed Load will compute by dividing the Total Load in these respect by the Total Length of

c) Since Loads from Wing Walls, Countrforts, Soil & Surcharge upon Well Cap are on Heel Side, thus Loads willalso considered as Uniformly Distributed Load but the Unit will be kN/m2. Value of these Uniformly DistribuedLoad will compute by dividing the Total Load in these respect by the Total Surface Area of Well Cap on Heel Side. Action of these Loads will be on Heel Side only.

d)will compute by dividing the Total Load in these respect by the Total Surface Area of Well Cap. These Load willbe active over the Wall Cap Total Surface Area.

e) For Calculation of Moment & Shear Total vertical Loads are considered be to an Uniformly Distributed Load over

compute by dividing the Total Vertical Loads upon Well Cap by the Total Surface Area of Well Cap. These Loadwill be active over the Total Surface Area of Well Cap.

ii)Soil & Surcharge :

a) Superstructure Dead Load Reaction (Without Utilites & Wearing Course) 1,976.278 kN

b) Superstructure Dead Load Reaction only for Utilites & WC 209.438 kN

c) 574.221 kN

d) Superstructure Live Load Reaction due to Lane Load 232.500 kN


Multiplier Factor for Earthquake -EQ gLL-EQ



Loads from Superstructure will act upon Well Cap through the Abutment Wall & its Position is in X-X Direction

resting on it will be Computated as Uniformly Distributed Load in X-X Direction having its Unit in kN/m. Value of

Well Cap in X-X Direction.

Self Weight of Well Cap will also be an Uniformly Distributed Load having its Unit in kN/m2. Value of these Load

the Total Surface Area of Well Cap for which the Unit of Pressure will kN/m2. The Value of these Load will be

Imposed Vertical Loads (LL+ DL) on Well Cap from Superstructure, Substructure, Self Weight of Well

RDL-Supr-1

RDL-Supr-2

Superstructure Live Load Reaction due to Wheel Load (Without IM) RLL-Supr-Wheel

RLL-Supr-Lane

e) Superstructure Live Load Reaction due to Pedestrian Load 112.500 kN

f) Dead Load Reaction from Abutment Wall Components (Stem+ 551.822 kNB-Wall+Seat).

d) Dead Load Reactions from Wing Wall Counterforts 184.325 kN

e) Dead Load Reactions from Wing Walls (Including Cantilevers) 366.494 kN

f) Dead Load Reaction from Self Weight of Well Cap 2,524.950 kN

g) Earth Pressure on Heel Side of Well Cap 2,102.265 kN

h) 297.264 kN

iii) Factored Vertical Load Reactions Upon Abutment Cap under Strength Limit State (USD) :

a) Factored Superstructure Dead Load (Without Utilites & Wearing Course) 2,470.348 kN

b) Factored Superstructure Dead Load Reaction only for Utilites & WC 314.156 kN

c) Factored Superstructure Live Load Reaction due to Wheel Load 1,336.500 kN


e) Factored Superstructure Live Load due to Pedestrian Load 196.875 kN

f) Factored Dead Load from Abutment Wall Components (Stem+ 689.778 kN

d) Factored Dead Load from Wing Wall Counterforts 230.407 kN

e) Factored Dead Load from Wing Walls (Including Cantilevers) 458.118 kN

f) Factored Dead Load due to Self Weight of Well Cap 3,156.188 kN

RLL-Supr-Pede.

RDL-Ab-Wall

RDL-WW-Countf

RDL-WW

RDL-Well-Cap-Self-Wt.

RDL-Soil.

Live Load Surcharge on Well Cap Heel Side (For H1 <6.00m) RLL-Sur.

= Dp-LL-Ab*WEq-Ab*H1*LAb-T-Inner

RDL-Supr-1-USD

= gDC*RDL-Supr-1

RDL-Supr-2-USD

= gDW*RDL-Supr-2

RLL-Supr-Wheel-USD

(Including IM) = m*IMgLL-TruckRLL-Supr-Wheel

RLL-Supr-Lane-USD

= mgLL-Lane*RLL-Supr-Lane

RLL-Supr-Pede.-USD

= mgLL-PL*RLL-Pede.

RDL-Ab-Wall-USD

B-Wall+Seat). = gDC*RDL-Ab-Wall

RDL-WW-Countf-USD

= gDC*RDL-WW-Countf

RDL-WW-USD

= gDC*RDL-WW

RDL-Self-Wt.-USD

= gDC*RDL-Well-Cap-Self-Wt.

g) Factored Earth Load on Heel Side of Well Cap 2,838.057 kN

h) 520.212 kN

i) Summasion all Factored Vertical Loads upon Well Cap 12,617.513 kN

iv) Computation of Factored Vertical Pressure on Well Cap through Abutment Wall for 1 (one) Meter StripLength in X-X Direction due DL & LL from Superstructure, Self Weight of Abutment & its Components :

a) Total Vertical Reaction on Well Cap from Superstructure, Self Weight 5,414.532 kN

b) 424.669 kN/mdue to Superstructure, Self Weight of Abutment & its Component.

v) Computation of Factored Vertical Pressure (DL & LL) on Well Cap due to Wing Walls, Soil & Surcharge

a) 37.931

b) 458.118 kN

c) 230.407 kN

d) 2,838.057 kN

e) 520.212 kN

f) Total Pressure upon Well Cap Heel Side due to Factored Loads from 4,084.724 kN Walls, Counterforts, Soil & Surcharge

g) Unit Pressure upon Well Cap on Heel Side from Wing Walls, Counterforts, 107.688

vi) Computation of Vertical Pressure (Load) on Well Cap due its Self Weight :

a) 3,156.188 kN

b) 47.192

vii) Computation of Total Vertical Pressure (Load) on Well Cap as Uniformly Distributed Load over Total

RDL-Soil.-USD

= gEV*RDL-Soil.

Factored Live Load Surcharge on Well Cap Heel Side (For H1 <6.00m) RLL-Sur.-USD

= gLL-LS.*RLL-Sur.

SRAll-V-Load.-USD

SRSupr+Ab-Wall-USD

of Abutment & its Components = (RDL-Supr-1-USD+ RDL-Supr-2-USD+ RLL-Supr-Wheel-USD

+ RLL-Supr-Lane-USD + RLL-Supr-Pede.-USD + RDL-Ab-Wall-USD)

Verticall Pressure on Well Cap for per meter Length in X-X Direction PSup+Ab-Wall-USD

= SRSup+Ab-Wall-USD/LW-Cap-Trans

on Heel Side for Unit Surface Area (kN/m2):

Surface Area of Well Cap on Heel Side = LW-Cap-Trans*WWell-Cap-Hell AWell-Cap-Heel m2

Pressure from Wing Walls upon Heel Side = RDL-WW-USD PDL-W-Wall-USD

Pressure from Counterforts upon Heel Side = RDL-WW-Count-USD PDL-Count-USD

Pressure from Soil upon Heel Side = RDL-Soil.-USD PDL-Soil.-USD

Pressure from Surcharge upon Heel Side = RDL-Sur.-USD PDL-Sur.-USD

SPWell-Cap-Heel.-USD

pHeel.-USD kN/m2

Soil & Surcharge = SPWell-Cap-Heel-USD./AWell-Cap-Heel.

Pressure due to Self Weight of Well Cap = RDL-Self-Wt.-USD PSelf-Wt.-USD

Unit Pressure upon Well Cap for Self Weight = PSelf-Wt-USD./AWell-Cap pSelf-Wt.-USD kN/m2

Surface Area of Well Cap :

a) 66.879

b) 12,617.513 kN

c) 188.661

viii) Calculations of Upward Reactions at Supports having Span in Y-Y Direction :

a) 4.900 m

b) 424.669 kN/mDirection due to Superstructure, Self Weight of Abutment & its Component.

c) Unit Pressure upon Well Cap on Heel Side from Wing Walls, Counterforts, 107.688 Soil & Surcharge.

d) Unit Pressure upon Well Cap for Self Weight 47.192

e)Span Beam of 1.000m Width.

424.6691 kN/m

107.6876

47.192

2.450 m 2.450 m

4.900

525.832 kN/m 393.915 kN/m

f) 525.832 kN/m

g) Reaction RB at Support Position-B on Toe End, 393.915 kN/m

ix) Moments in Y-Y Direction (about X-X) at different Points due to Factored (USD) Vertical Loads :

a) 823.455 kN-m/m

823.455*10^6 N-mm/m

b) 759.644 kN-m/m

Surface Area of Well Cap = AWell-Cap AWell-Cap m2

Total Vertical Loads upon Well Cap = SRAll-V-Load.-USD SRAll-V-Load.-USD

Vertical Load upon Well Cap for per meter Square per meter Length in X-X wX-X-USD kN/m2/m

Direction = SRAll-V-Load-USD./AWell-Cap

Span Length in Y-Y Direction (C/C Distance between Well Walls.) = SEff-Y-Y. LY-Y.

Concentrated Vertical Pressure on Well Cap for per meter Length in X-X PSup+Ab-Wall-USD

pHeel.-USD kN/m2/m

pSelf-Wt.-USD kN/m2/m

Sketch Diagram of Different Imposed Loads (Factored) & Reactions in Y-Y Direction as Simple Supported Single

PSup+Ab-Wall =

pHeel. = kN/m2/m

pSelf-Wt. = kN/m2/m

LY-Y/2 = LY-Y/2 =

LY-Y =

RA = RB =

Reaction RA at Support Position-A on Heel End, RA-USD

=((PSup+Ab-Wall-USD*LY-Y/2)+(pSelf-Wt.-USD*LY-Y2/2)+(PHeel-USD*LY-Y/2*(LY-Y/4+LY-Y/2)))/LY-Y

RB-USD

RB-USD =((PSup+Ab-Wall-USD*LY-Y/2)+(pSelf-Wt.-USD*LY-Y2/2)+(pHeel-USD*(LY-Y/2)2)/2)/LY-Y

(+) ve Bending Moment at Middle of Span at Position C, (+)MMidd-Y-Y-USD

MMidd-Y-Y-USD = (RB-USD*LY-Y/2 - pSelf-Wt.-USD*(LY-Y/2)2/2)

(+)ve Bending Moment Just on Outer Face of Abutment Wall on Toe (+)MAb-Outer-Toe.-USD

A C BHeel End Toe EndC

759.644*10^6 N-mm/m

c) 852.469 kN-m/m

852.469*10^6 N-mm/m

d) 143.810 kN-m/m

143.810*10^6 N-mm/m

e) Max. (+) Moment Value in Y-Y Direction occurs on Inner Face of 852.469 kN-m/mAbutment Wall on Heel Side of Well Cap. 852.469*10^6 N-mm/m

x) Shearing Force in Y-Y Direction at different Points due to Factored (USD) Vertical Loads :

i) Shearing Force at Inner Face of Well Wall on Heel Side. 479.368 kN/m

479.368*10^3 N/m

j) Shearing Force at just Inner Face of Abutment Wall on Heel Side. 111.528 kN/m

111.528*10^3 N/m

k) Shearing Force at Inner Face of Well Wall on Toe Side. 379.757 kN/m

379.757*10^3 N/m

l) Shearing Force at Span Center Position (278.293) kN/m

(-)278.293*10^3 N/m

m) Shearing Force at just Outer Face of Abutment Wall on Toe Side. 288.912 kN/m

288.912*10^3 N/m

xi) Calculations of Upward Reactions at Supports having Span in X-X Direction :

a) 4.050 m

b) Total Span Length of Well 12.150 m

c) 188.661

d)

188.661 kN/m

4.050 m 4.05 4.050 m

382.039 m 764.079 764.079 m 382.039 m

Side, MAb-Out-Toe.-USD = (RB-USD*(LY-Y/2-tAb-wal-Top./2) - pSelf-Wt.-USD*(LY-Y/2-tAb-wal-Top./2)2/2)

(+) ve Bending Moment Just on Inner Face of Abutment Wall on Heel (+)MAb-Inn-Heel.-USD

Side, MAb-Inn-Heel.-USD = RA-USD*(WWell-Cap-Hell - tWall./2)

- pSelf-Wt.-USD*(WWell-Cap-Hell - tWall./2)2/2- pHeel-USD*(WWell-Cap-Hell - tWall./2)2/2

(+) ve Bending Moment Just at Inner Face of Well Wall on Heel Side, (+)MHell-Well-Wall.-USD

MHeel-Well-Wall.-USD = RA-USD*tWall/2 - pSelf-Wt.-USD*(tWall/2)2/2 - pHeell-USD*(tWall/2)2/2

(+)MMax.-USD-Y-Y

VHeel-W-Wall.-USD

VHeel-W-Wall-USD = RA-USD - (pHeel.-USD+ pSelf-Wt.-USD)*tWall/2

VAb-Heel-USD

VAb-Heel-USD = RA-USD - (pHeel.-USD + pSelf-Wt.-USD)*(WWell-Cap-Hell - tWall./2)

VToe-W-Wall.-USD

VToe-W-Wall-USD = RB-USD - pSelf-Wt.-USD*tWall/2

VCenter-Span.-USD

VCenter-Span-USD = RA-USD - (pHeel.-USD+ pSelf-Wt-USD.)*LY-Y/2 - PSup+Ab-USD

VAb-Toe-USD

VAb-Toe-USD = RB-USD - (pSelf-Wt.-USD)*(LY-Y/2- tAb-wal-Top./2)

Span Length in X-X Direction Central & Side Pockets of Well (C/C Distance LX-X

between Well Walls) = SEffX-X

LX-X-Total

Vertical Load upon Well Cap for per meter Square in X-X Direction. wX-X kN/m2/m

Sketch Diagram of Imposed Loads (Factored) & Reactions in X-X Direction as a Multi Span Continuous Beam in X-X Direction with 1.000m Width.

wX-X =

LX-X = LX-X = m LX-X =

R1L = R2L = R2R = R1R =

12.150 m

e) 382.039 kN

f) 764.079 kN

xii) Moments in X-X Direction (about Y-Y) at different Points due to Factored (USD) Vertical Loads :

a) (+) ve Bending Moment at Middle Position of Each Span 110.519 kN-m/m

110.519*10^6 N-mm/m

b) 154.726 kN-m/m

154.726*10^6 N-mm/m

xiii) Shearing Force in X-X Direction at different Points due to Factored (USD) Vertical Loads :

a) 353.740 kN/m

353.74010^3 N/m

b) 353.740 kN/m

353.74010^3 N/m

x) Critical Section of Well Cap in Combined application of Shearing Force both in Y-Y & X-X Direction :

a) The Critical Section of Well Cap in respect of Shearing Force which has the 479.368 kN/mHighest Calculated Factored Shear Force. The Section on Inner Face of Well 492.097*10^3 N/m

9 Computation of Factored Moments & Shearing Forces under Service Limit State (WSD) :

i) Factored Vertical Load Reactions Upon Abutment Cap under Service Limit State (WSD) :

a) Factored Superstructure Dead Load (Without Utilites & Wearing Course) 1,976.278 kN

b) Factored Superstructure Dead Load Reaction only for Utilites & WC 209.438 kN

c) Factored Superstructure Live Load Reaction due to Wheel Load 574.221 kN


e) Factored Superstructure Live Load due to Pedestrian Load 112.500 kN

STotal =

Reaction at Support R1L = R1R = wX-X *LX-X/2 R1L& R1R

Reaction at Support R2L = R2R = 2*wX-X *LX-X/2 R2L& R2R

(+)MMidd-X-X-USD

MMidd-USD = wX-X*1/14*(LX-X)2/2

(-) ve Bending Moment at Support Positions R1 & R2, (-)MWall-R1&R2-X-X-USD

MR1&R2-X-X-USD = wX-X-USD*1/10*(LX-X)2/2

Shearing Force on Inner Face of Well Wall at Support Position R1,. VR1-X-X-USD

VR1-X-X-USD = wX-X-USD*(LX-X - tWall/2)/2

Shearing Force at both Face of Well Wall on Support Position R2,. VR2-X-X-USD

VR2-X-X-USD = wX-X-USD*(LX-X - tWall/2)/2

VCritical.

Wall on Heel Side in Y-Y Direction posses the Highest Shearing Force

RDL-Supr-1-WSD

= gDC*RDL-Supr-1

RDL-Supr-2-WSD

= gDW*RDL-Supr-2

RLL-Supr-Wheel-WSD

(Including IM) = m*IMgLL-TruckRLL-Supr-Wheel

RLL-Supr-Lane-WSD

= mgLL-Lane*RLL-Supr-Lane

RLL-Supr-Pede.-WSD

f) Factored Dead Load from Abutment Wall Components (Stem+ 551.822 kN

d) Factored Dead Load from Wing Wall Counterforts 184.325 kN

e) Factored Dead Load from Wing Walls (Including Cantilevers) 2,524.950 kN

f) Factored Dead Load due to Self Weight of Well Cap 2,524.950 kN

g) Factored Earth Load on Heel Side of Well Cap 2,102.265 kN

h) 297.264 kN

i) Summasion all Factored Vertical Loads upon Well Cap 11,290.514 kN

ii) Computation of Factored Vertical Pressure on Well Cap through Abutment Wall for 1 (one) Meter StripLength in X-X Direction due DL & LL from Superstructure, Self Weight of Abutment & its Components :

a) Total Vertical Reaction on Well Cap from Superstructure, Self Weight 3,656.759 kN

b) 286.805 kN/mdue to Superstructure, Self Weight of Abutment & its Component.

iii) Computation of Factored Vertical Pressure (DL & LL) on Well Cap due to Wing Walls, Soil & Surcharge

a) 37.931

b) 2,524.950 kN

c) 184.325 kN

d) 2,102.265 kN

e) 297.264 kN

= mgLL-PL*RLL-Pede.

RDL-Ab-Wall-WSD

B-Wall+Seat). = gDC*RDL-Ab-Wall

RDL-WW-Countf-WSD

= gDC*RDL-WW-Countf

RDL-WW-WSD

= gDC*RDL-WW

RDL-Self-Wt.-WSD

= gDC*RDL-Well-Cap-Self-Wt.

RDL-Soil.-WSD

= gEV*RDL-Soil.

Factored Live Load Surcharge on Well Cap Heel Side (For H1 <6.00m) RLL-Sur.-WSD

= gLL-LS.*RLL-Sur.

SRAll-V-Load.-WSD

SRSupr+Ab-Wall-WSD

of Abutment & its Components = (RDL-Supr-1-WSD+ RDL-Supr-2-WSD+ RLL-Supr-Wheel-WSD

+ RLL-Supr-Lane-WSD + RLL-Supr-Pede.-WSD + RDL-Ab-Wall-WSD)

Verticall Pressure on Well Cap for per meter Length in X-X Direction PSup+Ab-Wall-WSD

= SRSup+Ab-Wall-WSD/LW-Cap-Trans

on Heel Side for Unit Surface Area (kN/m2):

Surface Area of Well Cap on Heel Side = LW-Cap-Trans*WWell-Cap-Hell AWell-Cap-Heel m2

Pressure from Wing Walls upon Heel Side = RDL-WW-WSD PDL-W-Wall-WSD

Pressure from Counterforts upon Heel Side = RDL-WW-Count-WSD PDL-Count-WSD

Pressure from Soil upon Heel Side = RDL-Soil.-WSD PDL-Soil.-WSD

Pressure from Surcharge upon Heel Side = RDL-Sur.-WSD PDL-Sur.-WSD

f) Total Pressure upon Well Cap Heel Side due to Factored Loads from 5,146.735 kN Walls, Counterforts, Soil & Surcharge

g) Unit Pressure upon Well Cap on Heel Side from Wing Walls, Counterforts, 135.686

iv) Computation of Vertical Pressure (Load) on Well Cap due its Self Weight :

a) 2,524.950 kN

b) 37.754

v) Computation of Total Vertical Pressure (Load) on Well Cap as Uniformly Distributed Load over TotalSurface Area of Well Cap :

a) 66.879

b) 11,290.514 kN

c) 168.820

vi) Calculations of Upward Reactions at Supports having Span in X-X Direction :

a) 4.900 m

b) 286.805 kN/mDirection due to Superstructure, Self Weight of Abutment & its Component.

c) Unit Pressure upon Well Cap on Heel Side from Wing Walls, Counterforts, 135.686 Soil & Surcharge.

d) Unit Pressure upon Well Cap for Self Weight 37.754

e)Span Beam of 1.000m Width.

286.805 kN/m

135.6859

37.754

2.450 m 2.450 m

4.900

SPWell-Cap-Heel.-WSD

pHeel.-WSD kN/m2

Soil & Surcharge = SPWell-Cap-Heel-WSD/AWell-Cap-Heel.

Pressure due to Self Weight of Well Cap = RDL-Self-Wt.-WSD PSelf-Wt.-WSD

Unit Pressure upon Well Cap for Self Weight = PSelf-Wt-WSD./AWell-Cap pSelf-Wt.-WSD kN/m2

Surface Area of Well Cap = AWell-Cap AWell-Cap m2

Total Vertical Loads upon Well Cap = SRAll-V-Load.-WSD SRAll-V-Load.-WSD

Vertical Load upon Well Cap for per meter Square per meter Length in X-X wX-X-WSD kN/m2/m

Direction = SRAll-V-Load-WSD./AWell-Cap


Concentrated Vertical Pressure on Well Cap for per meter Length in X-X PSup+Ab-Wall-WSD

pHeel.-WSD kN/m2/m

pSelf-Wt.-WSD kN/m2/m

Sketch Diagram of Different Imposed Loads (Factored) & Reactions in Y-Y Direction as Simple Supported Single

PSup+Ab-Wall =

pHeel. = kN/m2/m

pSelf-Wt. = kN/m2/m

LY-Y/2 = LY-Y/2 =

LY-Y =

A C BHeel End Toe EndC

485.222 kN/m 319.007 kN/m

f) 485.222 kN/m

g) Reaction RB at Support Position-B on Toe End, 319.007 kN/m

vii) Moments in Y-Y Direction (about X-X) at different Points due to Factored (WSD) Vertical Loads :

a) 668.258 kN-m/m

668.258*10^6 N-mm/m

b) 668.258 kN-m/m

668.258*10^6 N-mm/m

c) 677.434 kN-m/m

677.434*10^6 N-mm/m

d) 129.957 kN-m/m

135.039*10^6 N-mm/m


viii) Shearing Force in Y-Y Direction at different Points due to Factored (WSD) Vertical Loads :

a) Shearing Force at Inner Face of Well Wall on Heel Side. 433.190 kN/m

433.190*10^3 N/m

b) Shearing Force at just Inner Face of Abutment Wall on Heel Side. 21.271 kN/m

21.271*10^3 N/m

c) Shearing Force at Inner Face of Well Wall on Toe Side. 307.681 kN/m

307.681*10^3 N/m

d) Shearing Force at Span Center Position (226.510) kN/m

(-)226.510*10^3 N/m

e) Shearing Force at just Outer Face of Abutment Wall on Toe Side. 235.005 kN/m

235.005*10^3 N/m

ix) Calculations of Upward Reactions at Supports having Span in X-X Direction :

a) 4.050 m

RA = RB =

Reaction RA at Support Position-A on Heel End, RA-WSD

=((PSup+Ab-Wall-WSD*LY-Y/2)+(pSelf-Wt.-WSD*LY-Y2/2)+(PHeel-WSD*LY-Y/2*(LY-Y/4+LY-Y/2)))/LY-Y

RB-WSD

RB-WSD =((PSup+Ab-Wall-WSD*LY-Y/2)+(pSelf-Wt.-WSD*LY-Y2/2)+(pHeel-WSD*(LY-Y/2)2)/2)/LY-Y

(+) ve Bending Moment at Middle of Span at Position C, (+)MMidd-WSD

MMidd-Y-Y-WSD = (RB-WSD*LY-Y/2 - pSelf-Wt.-WSD*(LY-Y/2)2/2)

(+)ve Bending Moment Just on Outer Face of Abutment Wall on Toe MAb-Outer-Toe.-WSD

Side, MAb-Out-Toe.-WSD = (RB-WSD*(LY-Y/2-tAb-wal-Top./2) - pSelf-Wt.-WSD*(LY-Y/2-tAb-wal-Top./2)2/2)

(+) ve Bending Moment Just on Inner Face of Abutment Wall on Heel (+)MAb-Inn-Heel.-WSD

Side, MAb-Inn-Heel.-USD = RA-WSD*(WWell-Cap-Hell - tWall./2)

- pSelf-Wt.-WSD*(WWell-Cap-Hell - tWall./2)2/2- pHeel-WSD*(WWell-Cap-Hell - tWall./2)2/2

(+) ve Bending Moment Just at Inner Face of Well Wall on Heel Side, (+)MHell-Well-Wall.WSD

MHeel-Well-Wall.WSD = RA-WSD*tWall/2 - pSelf-Wt.-WSD*(tWall/2)2/2 - pHeell-WSD*(tWall/2)2/2

(+)MMax.-WSD-Y-Y

VHeel-W-Wall.-WSD

VHeel-W-Wall-WSD = RA-WSD - (pHeel.-WSD+ pSelf-Wt.-WSD)*tWall/2

VAb-Heel-WSD

VAb-Heel-WSD = RA-WSD - (pHeel.-WSD + pSelf-Wt.-WSD)*(WWell-Cap-Hell - tWall./2)

VToe-W-Wall.-WSD

VToe-W-Wall-WSD = RB-WSD - pSelf-Wt.-WSD*tWall/2

VCenter-Span.-WSD

VCenter-Span-WSD = RA-WSD - (pHeel.-WSD+ pSelf-Wt-WSD.)*LY-Y/2 - PSup+Ab-WSD

VAb-Toe-WSD

VAb-Toe-WSD = RB-WSD - (pSelf-Wt.-WSD)*(LY-Y/2- tAb-wal-Top./2)

Span Length in X-X Direction Central & Side Pockets of Well (C/C Distance LX-X

b) Total Span Length of Well 12.150 m

c) 168.820

d)

168.820 kN/m

4.050 m 4.050 4.050 m

341.860 m 683.720 683.720 m 341.860 m

12.150 m

e) 341.860 kN

f) 683.720 kN

x) Moments in X-X Direction (about Y-Y) at different Points due to Factored (WSD) Vertical Loads :


98.895*10^6 N-mm/m

b) 138.453 kN-m/m

138.453*10^6 N-mm/m

xi) Shearing Force in X-X Direction at different Points due to Factored (WSD) Vertical Loads :

a) 316.537 kN/m

316.537*10^3 N/m

b) 316.537 kN/m

316.537*10^3 N/m

xii) Critical Section of Well Cap in Combined application of Shearing Force both in Y-Y & X-X Direction :

a) The Critical Section of Well Cap in respect of Shearing Force which has the 433.190 kN/mHighest Calculated Factored Shear Force. The Section on Inner Face of Well 433.190*10^3 N/m

10 Calculations of Unfactored Vertical Dead Loads on Abutment Well Cap & Moments :

i) Computation of Unfactord Dead Load (DL) Vertical Pressure on Well Cap from Different Components of Bridge Superstructure, Substructure (Abutment, Wing Walls& Counterforts, Well Cap) & Soil :

between Well Walls) = SEffX-X

LX-X-Total

Vertical Load upon Well Cap for per meter Square in X-X Direction. wX-X-WSD kN/m2/m

Sketch Diagram of Imposed Loads (Factored) & Reactions in X-X Direction as a Multi Span Continuous Beam in X-X Direction with 1.000m Width.

wX-XWSD =


R1 = R2 = R2 = R1 =

STotal =



(+)MMidd-X-X-WSD

MMidd-WSD = wX-XWSD*1/14*(LX-X)2/2

(-) ve Bending Moment at Support Positions R1 & R2, (-)MWall-R1&R2-X-X-WSD

MR1&R2-X-X-WSD = wX-X-WSD*1/10*(LX-X)2/2

Shearing Force on Inner Face of Well Wall at Support Position R1,. VR1-X-X-WSD

VR1-X-X-WSD = wX-X-WSD*(LX-X - tWall/2)/2

Shearing Force at both Face of Well Wall on Support Position R2,. VR2-X-X-WSD

VR2-X-X-WSD = wX-X-WSD*(LX-X - tWall/2)/2

VCritical.

Wall on Heel Side in Y-Y Direction posses the Highest Shearing Force

a) 37.931

b) Total Vertical Pressure on Well Cap from Superstructure, Self Weight 2,737.538 kN

c) 214.709 kN/mdue to Superstructure, Self Weight of Abutment & its Component.

d) 366.494 kN

e) 184.325 kN

f) 2,102.265 kN

g) 297.264 kN

h) 2,950.348 kN from Walls, Counterforts, Soil & Surcharge

i) 77.781

j) 2,524.950 kN

k) 37.754

l) 8,212.836 kN

m) 122.801

ii) Calculations of Unfactored Upward Dead Load Reactions at Supports having Span in X-X Direction :

a) 4.900 m

b)Single Span Beam of 1.000m Width.

214.709 kN/m

77.781

37.754

Surface Area of Well Cap on Heel Side = LAb-T-W-Cap*WWell-Cap-Hell AWell-Cap-Heel m2

SPDL-Supr+Ab-UF

of Abutment & its Components = (RDL-Supr-1+ RDL-Supr-2+ RDL-Ab-Wall)

Verticall Pressure on Well Cap for per meter Length in X-X Direction PDL-Sup+Ab-Wall-UF

= SPDL-Sup+Abl-UF/LAb-T-W-Cap

DL Pressure from Wing Walls upon Heel Side = RDL-WW-UF PDL-W-Wall-UF

DL Pressure from Counterforts upon Heel Side = RDL-WW-Count-UF PDL-Count-UF

DL Pressure from Soil upon on Heel Side = RDL-Soil.-UF PDL-Soil.-UF

DL Pressure from Surcharge upon Heel Side = RDL-Sur.-UF PDL-Sur.-UF

Total DL Pressure upon Well Cap Heel Side due to Factored Loads SPWell-Cap-Heel.-UF

Unfactored DL Unit Pressure upon Well Cap on Heel Side from Wing pUF-DL-Heel-UF. kN/m2

Wall Counterforts, Soil & Surcharge = SPWell-Cap-Heel.-UF/AWell-Cap-Heel.

DL Pressure due to Self Weight of Well Cap = RDL-Self-Wt.-UF PSelf-Wt.-UF

DL Unit Pressure upon Well Cap for Self Weight = PSelf-Wt-UF./AWell-Cap pDL-W-Cap-UF kN/m2

Summation of all Unfactored Vertical DL upon Well Cap = SRAll-V-DL.-UF SRAll-V-DL.-UF

Unfactored Vertical DL upon Well Cap for per meter Square per meter wX-X-UF kN/m2/m

Length in X-X Direction = SFRAll-V-DL-UF./AWell-Cap


Sketch Diagram of Different Imposed Unfactored Dead Loads & Reactions in Y-Y Direction as Simple Supported

PUF-DL-Sup+Ab-Wall =

pUF-DL-Heel.= kN/m2/m

pUF-DL-W-Cap = kN/m2/m

A C BHeel End Toe End

2.450 m 2.450 m

4.9 m

342.775 kN/m 247.493 kN/m

c) 342.775 kN/m

d) 247.493 kN/m

iii) Unfactored Moments in Y-Y Direction (about X-X) at different Points due to Vertical Dead Loads :

a) 493.048 kN-m/m

493.048*10^6 N-mm/m

b) 457.218 kN-m/m

457.218*10^6 N-mm/m

c) 503.559 kN-m/m

503.559*10^6 N-mm/m

d) 92.434 kN-m/m

95.489*10^6 N-mm/m


iv) Calculations of Unfactored Upward Dead Load Reactions at Supports having Span in Y-Y Direction :

a)

122.801 kN/m

4.050 m 4.050 4.050 m

248.672 m 497.345 497.345 m 248.672 m

12.150 m

e) 248.672 kN

f) 497.345 kN

v) Unfactored Moments in X-X Direction (about Y-Y) at different Points due to Vertical Dead Loads :

LY-Y/2 = LY-Y/2 =

LY-Y =

RAUF = RBUF =

Reaction RA at Support Position-A on Heel End, RA-UF

=((PUF-DL-Sup+Ab-Walll*LY-Y/2)+(pUF-DL-W-Cap.*LY-Y2/2)+(PUF-DL-Heel*LY-Y/2*(LY-Y/4+LY-Y/2)))/LY-Y

Reaction RB at Support Position-B on Toe End, RB-UF

RB-UF =((PDL-Sup+Ab-Wall-UF*LY-Y/2)+(pDL-W-Cap-UF*LY-Y2/2)+(pDL-Heel.-UF*(LY-Y/2)2)/2)/LY-Y

(+) ve Bending Moment at Middle of Span at Position C, (+)MDL-Midd-UF

MDL-Midd-Y-Y -UF = (RB-UF*LY-Y/2 - pDL-W-Cap-UF.*(LY-Y/2)2/2)

(+)ve Bending Moment Just on Outer Face of Abutment Wall on Toe (+)MAb-Outer-Toe.-UF

Side, MDL-Ab-Out-Toe.-UF=(RB-UF*(LY-Y/2-tAb-wal-Top./2) - pDL-W-Cap.-UF*(LY-Y/2-tAb-wal-Top./2)2/2)

(+) ve Bending Moment Just on Inner Face of Abutment Wall on Heel (+)MAb-Inn-Heel.-UF

Side, MAb-Inn-Heel.-UF = RA*(WWell-Cap-Hell - tWall./2)

- pDL-W-Cap-UF*(WWell-Cap-Hell - tWall./2)2/2- pDL-Heel-UF.*(WWell-Cap-Hell - tWall./2)2/2

(+) ve Bending Moment Just at Inner Face of Well Wall on Heel Side, (+)MHell-Well-Wall.-UF

MHeel-Well-Wall.-UF = RA-UF*tWall/2 - pDL-W-Cap-UF*(tWall/2)2/2 - pDL-Heel-UF.*(tWall/2)2/2

(+)MMax.-UF-Y-Y

Sketch Diagram of Imposed Unfactored Dead Loads & Reactions in X-X Direction as a Multi Span Continuous Beam in X-X Direction with 1.000m Width.

wX-X-UF =


R1L = R2L = R2R = R1R =

STotal =




71.937*10^6 N-mm/m

b) 100.712 kN-m/m

100.712*10^6 N-mm/m

11 Features related to Flexural Design of Reinforcements for Well Cap in botn Y-Y & X-X Directions:

i) Design Strip Width for Well Cap in Y-Y & X-X Directions & Clear Cover of Different Faces;

a) b 1.000 m

b) 75 mm

75 mm

75 mm



b) c Variable

c) Variable

Variable

Variable

410.00

Variable

Variable mm

Variable mm




(+)MMidd-X-X-UF

MMidd-UF = wX-X-UF*1/14*(LX-X)2/2

(-) ve Bending Moment at Support Positions R1 & R2, (-)MWall-R1&R2-X-X-UF

MWall-1&R2-X-X -UF= wX-X-UF*1/10*(LX-X)2/2

Let Consider the Design Width in both Y-Y & X-X Directions are = 1000mm

Let the Clear Cover at Bottom Surface of Well Cap, C-Cov.Bot. = 75mm, C-Cov-Bot.

Let the Clear Cover at Top Surface of Well Cap, C-Cov.Top = 75mm, C-Cov-Top.

Let the Clear Cover at Vertical Surface of Well Cap, C-Cov.Vert.. =75mm, C-Cov-Side.

c/de-Max.











Centroid in mm.




thus de = ds .





Variable N-mm

Variable

0.240

240.000/10^3

240.000*10^6

2.887022688

c) 692885445.077 N-mm

692.89 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)

Location Value of Value of Actuat Acceptable M Maximum

of Section Unfactored Cracking Factored Allowable Flexuralunder Dead Load As per Moment Cracking Cracking Moment Factored Min. Moment Moment

for Moment Equation Value Moment Moment of Section Moment

Moment 5.7.3.3.2-1 M (1.33*M)Calculation kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

Mid. Span 493.048 692.885 692.885 692.885 831.463 823.455 1095.195 831.463 831.463

Ab-Out-Toe. 457.218 692.885 692.885 692.885 831.463 759.644 1010.327 831.463 831.463

Ab-In-Heel 503.559 692.885 692.885 692.885 831.463 852.469 1133.783 831.463 852.469

Heel-Wall 92.434 692.885 692.885 692.885 831.463 143.810 191.268 831.463 831.463

71.937 692.885 692.885 692.885 831.463 110.519 146.990 831.463 831.463

100.712 692.885 692.885 692.885 831.463 154.726 205.785 831.463 831.463












Snc = (b*hWell-Cap.3/12)/(hWell-Cap./2)










for RCC Mu


(+)ve. Monents in Y-Y Direction.

(+) ve. & (-) ve. Moments in X-X Direction.

(+) Middle

(-) on Wall

iv)


b) 0.016

12 Flexural Design of Reinforcements for Well Cap in Y-Y Directions:

i) Design of Reinforcements in Y-Y Direction (Parallel to Traffic) against Max. (+) ve Moment:

a) 823.455 kN-m/m

823.455*10^6 N-mm/m

831.463 kN-m/mReinforcements will be on Bottom Surface of Well Cap. 831.463*10^6 N-mm/m

b) 831.463 kN-m/m

831.463*10^6 N-mm/m

c) 25 mmDirection of Well Cap.

d) 490.874

e) The provided Effective Depth for the Section with Main Reinforcement on 1,112.500 mm

f) 42.689 mm

g) 2,045.157

h) 240.018 mm,C/C

i) 125 mm,C/C

j) 3,926.991

k) 0.004

i) 90.200 mm


pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


In Y-Y Direction the Maximum Moment occurs at Inner Face of Abutment (+)MMax.-USD-Y-Y

which is Less than Mr the Allowable Minimum Moment for Flexural Design.

Thus Mr is the Governing Moment for Flexural Design. For (+) ve, Moment the Mr

Since MMax.-USD-Y-Y > Mr, the Allowable Minimum Moment for the Section, thus MU

MMax.-USD-Y-Y is the Design Moment MU.

Let provide 25f bars as Main Reinforcement on Bottom Surface in Y-Y DMain-bar-Y-Y.

X-Sectional of 25f Main bars = p*DMain-bar2/4 Af-25. mm2

de-pro.

Bottom Surface, dpro = (hWell-Cap.-CCov-Bot. -DMain-bar-Y-Y./2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Bot.-Y-Y mm2/m

Spacing of Reinforcement with 25f bars = Af-25.b/As-req-Bot.-Y-Y sreq


spro = 125mm,C/C

The provided Steel Area with 25f bars having Spacing 150mm,C/C As-pro-Bot-Y-Y. mm2/m

= Af-25..b/spro



j) Resisting Moment for the Section with provided Steel Area, 1,718.585 kN-m/m

k) Mpro>Mu OK

l) ppro<pmax OK

m) Checking according to Provisions of AASHTO-LRFD-5.7.3.3.1 :

0.450

c 76.670 mm

0.85

0.069


ii) Checking Against Max. Shear Force on Well Cap in X-X Direction.

a) The Maximum Shear Force occurs at Inner Face of Well Wall on Heel Side, 479.368 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 479.368*10^3 N/m

b)

b-i) 1,000.000 mm

a-ii) 1,001.250 the neutral axis between Resultants of the Tensile & Compressive Forces due

1,001.250 mm 864.000 mm

b-iii) f 0.90

b-iv) - N

c) 5,256.563 kN/m 5256.563*10^3 N/m

Mpro




i) Accodring to AASHTO-LRFD-.7.3.3.1; In Flexural Design c/de £ 0.42; where, c/de-Max.

ii) c is the Distance between Neutral Axis& the Extrime Compressive Face,


ii) b1 is Factor for Rectangular Stress Block for Flexural Design b1

iii) Thus for the Section the Ratio c/de = 0.069 c/de-pro


VU.

Force, VMax = VHeel-W-Wall-USD.= VU






h = hWell-Cap Depth of Well Cap.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h




The Nominal Shear Resitance Vn for the Section is the Lesser value of any of Vn-Heel Equations as mentioned in Aritical 5.8.3.3 :

7,616.584 kN/m 7616.584*10^3 N/m

5,256.563 kN/m 5256.563*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)thus the Well does not require any Shear Reinforcement.

f)Section does not Require any Shear Reinforcement, thus the Flexural Design of Reinforcement on Bottom

iii) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 1,546.726 N-mm 1298.015*10^6 kN-m

1,718.585 N-mm 1442.239*10^6 kN-m

f 0.90

b)


d) Since Well Cap is being considered as s Simple Supported Rectangular Beam 1,718.585 kN-mhaving 1.000 m Wide Strips. The Steel Area against Factored Max. Moments 1442.239*10^6 N-mm













Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Well Cap

Surface of Well Cap in Y-Y Direction (Parallel to Traffic Direction) is OK.






y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-Y-Y

e) 823.455 kN-m

823.455*10^6 N-mm

f) Mr>MMax.-USD-Y-Y Satisfied

iv) Checking in respect of Control of Cracking By Distribution of Reinforcement, (AASHTO-LRFD-5.7.3.4) :

a)

Where;

b) 155.063

677.434 kN-m 677.434*10^6 N-mm

3,926.991


c) 171.349




17,000.000 N/mm

at its Mid Span will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment MU at Mid Span of assumed Rectangular MMax.-USD-Y-Y

Beam in Y-Y Direction = MMax.-USD-Y-Y


Factored Moment M for the Section ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State MMax.-WSD-Y-Y





i) dc=Depth of Concrete Extreme Tension Face from the Center of the Closest dc


of Rectangular Section, dc = DBar/2 + CCov-Bot. Since Clear Cover at Bottom of

Well Cap, CCover-Bot = 75mm & Bar Dia, DBar = 25f ; thus dc = (25/2 + 50)mm


Cover = 50mm.In Well Cap the Tension Bars in One Layer & as per Condition




Since the Structure will be a Buried Components thus the Allowable Max.

246.000

d)

e) 15,384.168 N/mm




i)

j)

13

i)

a) 110.519 kN-m/m

110.519*10^6 N-mm/m

831.463 kN-m/m 831.463*10^6 N-mm/m

Direction.

b) 831.463 kN-m/m

831.463*10^6 N-mm/m

c) 25.000 mmWell Cap.

d) 490.874

e) 1,087.500 mm


value is ZMax. = 17000N/mm









the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Y-Y Direction at Well Cap Bottom Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.

More over though the Structure is a Nonprestressed one & value of dc have not Exceeds 900 mm, thus Component does require any additional Longitudinal Skein Reinforcement.

Flexural Design of Reinforcements for Well Cap in X-X Directions for (+) Moment on Bottom Surface :

Design of Reinforcements in X-X Direction (Perpendicular to Traffic) against (+) ve Moment:

The Max. (+) ve Moment occurs just at Middle of all Spans in X-X (+)MMidd-X-X-USD

Direction the value of MMidd.-X-X-USD < Mr, Minimum Flexural Strength Moment

for the Section, thus Mr is Governing Moment for Design. For (+) ve Moment Mr

value the Reinforcements will be on the Bottom Surface of Well Cap in X-X

Since (-)MMidd-X-X-USD > Mr, the Allowable Minimum Moment for the Section, MU

thus (-) MMidd.-X-X-USD is the Design Moment MU.

Let provide 25f bars as Reinforcement on Bottom Surface in X-X Direction of DBar-X-X.-Bot

X-Sectional of 25f Bars = p*DBar-X-X-Bot.2/4 Af-25. mm2

The provided Effective Depth for the Section with Main Reinforcement in Y-Y on dpro.

f) 43.711 mm

g) 2,114.482

h) 232.149 mm, C/C

i) 200.000 mm,C/C

j) 2,454.369

k) 0.002

l) 56.375 mm

m) Resisting Moment for the Section with provided Steel Area,

1,705.563 kN-m/m

n) Mpro>Mu Not OK

o) ppro<pmax Not OK


a) 0.450

b) c 47.919 mm

c) 0.85

d) 0.044


iii) Checking Against Max. Shear Force on Well Cap in X-X Direction.

a) The Maximum Shear Force occurs at Inner Face of Well Wall on X-X Direction, 353.740 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 353.74010^3 N/m

Bottom Surface 25f, dpro = (hWell-Cap. - CCov-Bot. -DMain-Y-Y - DBarX-X.-Bot/2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Bot.-X-X mm2/m


Let the provided Spacing of Reinforcement with 25f bars for the Section spro.-X-X-Bot.

spro = 200mm,C/C

The provided Steel Area with 25f bars having Spacing 200mm,C/C As-pro-Bot-X-X. mm2/m

= Af-25.b/spro



= As-pro*fy(d - apro/2)/10^6 Mpro









VU.

Force, VMax = VR1-X-X-USD= VU

b)

b-i) 1,000.000 mm


978.750 mm 864.000 mm

b-iii) f 0.90

b-iv) - N

c) 3,164.306 kN/m 3164.306*10^3 N/m

3,164.306 kN/m 3164.306*10^3 N/m

5,138.438 kN/m 5138.438*10^3 N/m


c-ii) 0.000 N/m

d) Vn>Vu Satisfied


f)

iv) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 959.379 N-mm 959.379*10^6 kN-m










The Nominal Shear Resitance Vn for the Section is the Lesser value of any of Vn-X-X-Mid Equations as mentioned in Aritical 5.8.3.3 :










Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Shear Forcesfor the Section in X-X Direction > the Critical Shear Forces in Y-Y Direction, thus the Flexural Design of Bottom Surface Reinforcement of Well Cap in X-X Direction (Perpendicular to Traffic) is OK.


1,065.977 N-mm 1065.977*10^6 kN-m

f 0.90

b)


d) Since Well Cap is being considered as s Simple Supported Rectangular Beam 1,065.977 kN-mhaving 1.000 m Wide Strips. The Steel Area against Factored (+) Moments in 1065.977*10^6 N-mm

e) 110.519 kN-m

110.519*10^6 N-mm

f) Mr>MMidd-X-X-USD Satisfied

v) Checking in respect of Control of Cracking By Distribution of Reinforcement, (AASHTO-LRFD-5.7.3.4) :

a)

Where;

b) 37.052

98.895 kN-m 98.895*10^6 N-mm

2,454.369


c) 170.000






y(d/s-a/2)

Mn = Asfy(ds-a/2)

(+)Mn-X-X

X-X at its Mid Span will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment (+) M at Mid Span of assumed Rectangular (+)MMidd-X-X-USD

Beam in X-X Direction = (+) MMidd-X-X-USD


Factored Moment M for the Section ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





i) M is Calculated Moment for the Section under Service Limit State (+)MMidd-X-X-USD









17,000.000 N/mm


246.000

d)

e) 3,705.150 N/mm




i)

j)

14

i)

a) 154.726 kN-m/m




















Flexural Design of Reinforcements for Well Cap in X-X Directions for (-) Moment on Top Surface :

Design of Reinforcements in X-X Direction (Perpendicular to Traffic) against (-) ve Moment:

The Calculated (-)ve Moments occur at Support Positions of Well (-)MWall-R1&R2-X-X-USD

154.726*10^6 N-mm/m

831.463 kN-m/m(-) ve Moment value the Reinforcement will be on Top Surface of Well Cap in 831.463*10^6 N-mm/m

b) 831.463 kN-m/m

831.463*10^6 N-mm/m

b) 25.000 mmWell Cap.

c) 490.874

e) The provided Effective Depth for the Section with Main Reinforcement on 1,112.500 mm

f) 42.689 mm

g) 2,065.046

h) 237.706 mm,C/C

i) 200.000 mm,C/C

j) 2,454.369

k) 0.002

l) 56.375 mm

m) Resisting Moment for the Section with provided Steel Area, 1,091.134 kN-m/m

n) Mpro>Mu OK

o) ppro<pmax OK


a) 0.450

b) c 47.919 mm

Cap in X-X Direction having equal value. (-) MWall-R1&R2-X-X-USD < Mr, the

Required Minimum Flexural Strength Moment which Governs the Design. For Mr

X-X Direction.

Since (-) MWall-R1&R2-X-X-USD > Mr, the Allowable Minimum Moment for the MU

Section, thus (-) MWall-R1&R2-X-X-USD is the Design Moment MU.

Let provide 25f bars as Reinforcement on Top Surface in X-X Direction of DBar-X-X.-Top

X-Sectional of 25f Bars = p*DBar-X-X-Bot.2/4 Af-25. mm2

dpro.

Top Surface, dpro = (hWell-Cap. - CCov-Top - DBarX-X.-Top/2)



2))(1/2))

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-req-Top.-X-X mm2/m



spro = 200mm,C/C

The provided Steel Area with 25f bars having Spacing 200mm,C/C As-pro-Top-X-X. mm2/m

= Af-25b/spro



Mpro






c) 0.85

d) 0.043


iii) Checking Against Max. Shear Force on Well Cap in X-X Direction.

a) The Maximum Shear Force occurs at Inner Face of Well Wall on X-X Direction, 316.537 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 316.537*10^3 N/m

b)

b-i) 1,000.000 mm

a-ii) 1,001.250 the neutral axis between Resultants of the Tensile & Compressive Forces due

1,001.250 mm 864.000 mm

b-iii) f 0.90

b-iv) - N

c) 3,237.048 kN/m 3237.048*10^3 N/m

3,237.048 kN/m 3237.048*10^3 N/m

5,256.563 kN/m 5256.563*10^3 N/m


c-ii) 0.000 N/m





VU.

Force, VMax = VR1-X-X-USD= VU










The Nominal Shear Resitance Vn for the Section is the Lesser value of Vn-X-X-R1&R2 any of Equations as mentioned in Aritical 5.8.3.3 :







d) Vn>Vu Satisfied


f)

iv) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 956.493 N-mm 956.493*10^6 kN-m

1,062.770 N-mm 1062.770*10^6 kN-m

f 0.90

b)


d) Since Well Cap is being considered as s Simple Supported Rectangular Beam 1,062.770 kN-mhaving 1.000 m Wide Strips. The Steel Area against Factored (-) Moments in 1062.770*10^6 N-mm

e) 154.726 kN-m

154.726*10^6 N-mm

f) Mr>MWall-R1&R2-X-X-USD Satisfied


a)

Where;

b) 50.706




Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Shear Forcesfor the Section in X-X Direction > the Critical Shear Forces in Y-Y Direction, thus the Flexural Design of Bottom Surface Reinforcement of Well Cap in X-X Direction (Perpendicular to Traffic) is OK.






y(d/s-a/2)

Mn = Asfy(ds-a/2)

(-)Mn-X-X

X-X Well Wall Face will have value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored Moment (-) M at Well Wall Face of assumed (-)MWall-R1&R2-X-X-USD

Rectangular Beam in X-X Direction = (-) MWall-R1&R2-X-X-USD

Relation between the Computed Factored Flexural Resistance Mr & the

Actual Factored Moment M for the Section ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)





138.453 kN-m 138.453*10^6 N-mm

2,454.369


c) 170.000




17,000.000 N/mm


246.000

d)

e) 5,070.644 N/mm




i) M is Calculated Moment for the Section under Service Limit State (-)MWall-R1&R2-X-X-USD























i)

j)

15 Arrangement of Reinforcement on Top & Bottom Surface of Well Cap under Provision of Distribution Reinforcement or as Temperature & Shrinkage Reinforcement in Y-Y (Parallel to Traffic) & also X-X (Perpendicular to Traffic) Directions Including the Vertical Faces.

i) Arrangement of Reinforcement on Horizontal Surfaces :

a)

b) Let consider a 1.000m X 1.000m Strip of Surface of Well Cap for Calculation of 1.000 m

1.000 m

c) 256.250 Temperature & Shrinkage Reinforcement on any Cocrete Exposed Surface in

d) 268.293 Temperature Reinforcement for Structural Components having its Thickness

e) 1000000.000

f) 16.000 mmDirection.

g) 201.062

h) 749.413

i) Spacing of Shrinkage & Temperature Reinforcements will less of 3-times 300.000 mmthe Component thickness or 450mm. Since the Thickness of Component is1200mm; thus let provide Specing of Shrinkage & Temperature Reinforcements

j)




Since on Bottom Surface of Well Cap, Reinforcements are being Provided both in Y-Y & X-X Directions, but on Top Surface the Provided Reinforcements are in X-X Direction only. Thus it requires provision of Reinforcements in Y-Y Direction. On Top Surface in Y-Y Direction (Parallel to Traffic) Reinforcements can Provide as Shrinkage &Temperature Reinforcement on same Surface according to provisions of AASHTO-LRFD.- 5.10.8.

bY-Y.

Shrinkage & Temperature Reinforcement on Top Surface in Y-Y Direction. bX-X

Under Provision of AASHTO1996-8.20.1&.2. the Required Steel Area against AS-S&T mm2/m

both Directions = (1/8)*25^2*3.28mm2/m

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T-Top.Y-Y mm2

1200mm or Less; As ³ 0.11Ag/fy in both way.(Here Well Cap Thickness = 1200mm).

Here Ag is Gross Area of Strip on Top Surface = bY-Y*bX-X Ag-Strip mm2

Let provide 16f bars as Shrinkage & Temperature on Top of Well Cap in Y-Y DBar.S&T-Y-Y


Spacing of 16f bars as Shrinkage & Temperature on Top of Well Cap in sreq-S&T-Top-Y-Y. mm2

Y-Y Direction = Af-16.bX-X/As-req-S&T-Top.Y-Y

spro-S&T-Top-Y-Y.

for Top Surface in Y-Y Direction = 300m C/C.

Since As-pro-Top. > As-req-Top & spro-Top> s-req-Top, thus the provision of Shrinkage & Temperature on Top Surface

ii) Arrangement of Reinforcements on Vertical Faces both in Vertical & Horizontal Direction :

a) The Vertical Faces of Well Cap Require Reinforcements in both Vertical & Horizontal Directions, those can also

b) Since the Well Cap has a Thickness/Depth of 1200mm, thus let consider a 1.000 m

1.200 m

Shrinkage & Temperature Reinforcements on both Directions.

c) 321.951 Temperature Reinforcement for Structural Componentshaving its Thickness

d) 1200000.000

e) 16.000 mmin Horihontal Direction.

f) 201.062

g) 1.601 nos.

h) Let provide 4 nos 16f Bars as Shrinkage & Temperature on Vertical Faces in 4.000 nos.Horihontal Direction.

i) 400.000 mm

h) Spacing of Shrinkage & Temperature Reinforcements will less of 3-times 300.000 mmthe Component thickness or 450mm. Since the Thickness of Component is1200mm; thus let provide Specing of Shrinkage & Temperature Reinforcements

j)

k) Since the Well Cap Bottom & Top Surfaces are being Provided with Flexural as well Shrinkage & Temperature Reinforcement Bars having Spacing ranging 150mm to 300mm C/C, thus it is Recommended to bent up & down those Bars alternately from Bottom & Top. These arrangement will fulfill the requirements of Vertical Shrinkage & Temperature Reinforcements for Vertical Faces of Well Cap.

16 Checking against Shearing Forces at Critical Sections of Well Cap.

of Wel Cap in Y-Y Direction is OK.

provide as Shrinkage & Temperature Reinforcement under Provisions of AASHTO-LRFD.- 5.10.8.

LHor.

Strip of Well Cap Vertical Surface having Horizontal Length LHor. = 1000mm hVer.

& Vertical Height hVer. = 1200mm (Depth of Well Cap) for Calculations of its

According to AASHTO-LRFD-5.10.8.1. Steel Area required as Shrinkage As-req-S&T-V-Face mm2

1200mm or Less; As ³ 0.11Ag/fy in both way.(Here Well Cap Thickness = 1200mm).

Here Ag is Gross Area of Strip on Vertical Surface = LHor.*hVer. Ag-Strip mm2

Let provide 16f bars as Shrinkage & Temperature on Well Cap Vertical Faces DBar-Hor

X-Sectional Area of 16f bar = pDBar-Hor2/4 Af-16 mm2

Number of 16f Bars required as Shrinkage & Temperature on Vertical Faces NBar-Hor-req.

in Horihontal Direction = As-req-S&T-V-Face/Af-16.

NBar-Hor-pro.

Spacing of 4nos. 16f bars as Shrinkage & Temperature on Well Cap Vertical spro-S&T-Hor.

Faces in Horizopntal Direction = hVer./(NBar-Hor-pro.- 1)

spro-S&T-Top-Y-Y.

for Top Surface in Y-Y Direction = 300m C/C.

Since As-pro-Top. > As-req-Top & spro-Top> s-req-Top, thus the provision of Shrinkage & Temperature on Top Surfaceof Wel Cap in Y-Y Direction is OK.

i) Checking of Shearing Forces at Critical Sections of Well Cap along X-X Direction due to Vertical Loads

a)

Back Face of Abutment Well on Heel Side (Earth Face).

b) Since the Well Cap is being Designed Considering it has One Way Action; thus the Shear Resistances for both the

c) The Calculated Factored Shearing Forces on Toe Side of Well Cap at Critical 348.492 kN/m

d) 5,256.563 kN/m 52563.563*10^3 N/m

7,616.584 kN/m 7616.584*10^3 N/m

5,256.563 kN/m 52563.563*10^3 N/m

d-i) 7,616.584 kN/m(AASHTO-LRFD- Equ. 5.8.3.3-1); 7616.584*10^3 N/m

d-ii) - N/m

d-iii) b 2.000

d-iv) - N

e) Vn>Vu Satisfied

f)thus the Well does not require any Shear Reinforcement.

g)

ii) Checking of Punching Shear Forces at Critical Sections of Well Cap along X-X Direction due to Vertical

from Superstructure (DL + LL), Loads of Abutment Wall (DL), Soil & Surcharge Loads (DL + LL).

According to AASHTO-LRFD-5.8.3.2; 5.13.3.6.1 & C 5.13.3.6.1 the Critical Sections of Footings (Here Well Cap)

prevaile on 2 (Two) Loactions; i) At a Distance dv from Face of Abutment Wall Toe Side (River Face), ii) Just on the

Locations should Satisfy the Provisions of AASHTO-LRFD-5.8.3.

VU-dv-Toe

Section at a Distance dv from the = RB-USD - (LY-Y/2 - tAb-Wal-Bot/2 - dv)*pSelf-Wt.-USD

The Nominal Shear Resitance Vn for the Section is the Lesser value of any of Vn-Heel Equations as mentioned in Aritical 5.8.3.3 :






For Footing/Foundation Slab Vs = 0.

b is Factor for the Diagonally Cracked Concrete to transmit Tension as per AASHTO-LRFD-5.8.3.4. For Footing/Foundation Slab b = 2.00.






Checking of Shearing Forces at Critical Section on Abutment Wall Heel Face of Well Cap along X-X Direction due to Vertical Loads from Superstructure (DL + LL), Loads of Abutment Wall (DL), Soil & Surcharge Loads (DL + LL) have already been done in Serial No-12-ii & found Satisfactory, thus no further Checking ie Required.

Loads from Superstructure (DL + LL), Loads of Abutment Wall (DL), Soil & Surcharge Loads (DL + LL) as

Two-Way Action According to Provision of AASHTO-LRFD-5.13.3.6 :

a)

b)

c) 1.864

d) 26.005 m

26.005*10^3 mm

e) 1,001.250 mm

f) The Calculated Factored Total Shearing Forces at Critical on Toe Side at 4,075.105 kN

g) The Calculated Factored Total Shearing Forces at Critical on Heel Side at 7,130.067 kN

h) 41,412.369 kN

i) 180439.9183 kN180439.918*10^3 N

j) Vn1<Vn2 Satisfied

k)Safe in Respect Critical Shear.

l)

m)

According to AASHTO-LRFD-5.13.3.6.1 the Critical is Located at a Distance not Less than 0.5dv to Calcutate the

Parimeter bo of the Concetrated Loads.

According to AASHTO-LRFD-5.13.3.6.3 for Two-Way Action without Transverse/Shear Reinforcement the Nominal

Shear Resitance of Concrete for the Section, Vn = (0.17+0.33/bc)Öf/cbodv £ 0.33 f/

cbodv ; where,

bc is Ratio of Long Side to Short Side of Rectangule through which the Horces bc

Concentrated Load or Reaction Transmitted having value = LAb-Wall-Trans./LAb-Cap-Long.

bo is Perimeter of the Critial Section at a distance 0.5*dv from Face of Load bo

Action Face having value, bo = 2*{(2*0.5dv +LAB-Wall-Trans) + (tAb-Wall-Bot+2*0.5dv)} /103m

dv is Effective Shear Depth in mm. dv

VU-Total-Crit-Toe-1

a Distance 0.5*dv from the Abutment Wall Face

= ((RB-USD - (LY-Y/2 - tAb-Wal-Bot/2 - 0.5*dv)*pSelf-Wt.-USD)*LW-Cap-Trans.

VU-Total-Crit-Toe-2

a Distance 0.5*dv from the Abutment Wall Face

= ((RA-USD - (LY-Y/2 - tAb-Wal-Bot/2 - 0.5*dv)*pSelf-Wt.-USD+ pHeel)*LW-Cap-Trans.

Computed value on Toe Side for Vn-1 = (0.17+0.33/bc)Öf/cbodv Vn-1

Computed value on Toe Side for Vn-2 = 0.33 f/cbodv Vn-2

Relation betweenComputed values of Vn-1 & Vn-2 for Tor Side

All Provisions of AASHTO-LRFD-5.13.3.6.1 & AASHTO-LRFD-5.13.3.6.3 are being Satisfied, thus the Structure is

Since VC-d2 > VC-d1 > VU, Thus the Well Cap does not Require any Punching Shear Reinforcement.

Since the Resisting Moment > Designed Moment, the Provided Steel Ratio < Max. Steel Ratio, the Calculated Imposed Max. Shear Forces (Nominal & Punching) < the Computed Shear Forces (Nominal & Punching) and the Well Cap does not require any Shear Reinforcement, thus the Flexural Design of the Well Cap for Reinforcement on Top & Bottom Surface in all respect (both in X-X & Y-Y Direction) is OK.

Maximum

Mr)

from Face of Abutment Wall Toe Side (River Face), ii) Just on the


P. Structural Design of RCC Wells for Bridge foundation:

1

2 Dimensions of Bridge Superstructur, Substructure & RCC Well for Foundation :












C

C

SL

SAddl.

LGir.

WCarr-Way.

WS-Walk.

WCurb.

WR-Post.

WB-Deck.

25251775

1900

2147

600

300

30 0

700

4300

7503000 450

450

1200

600

5225

1200

2000

6150

AB

H1

=

4947 H =

61

47

1500

1447

5500

600

10250

9350

12750

3450 3450 3450600 600 600600

34503450

60 0

2750

60 0

2525

1775

5500

3000

450

450

450

450

2150

2150

2450

2750

RL-5.00m

RL-2.20m

3000 2100

300
















w) C/C Distance Between Main Girders 2.000 m

w-i) Distance of Slab Outer Edge to Exterior Girder Center 1.125 m







ii) Dimensions of Sub-Structure.







g) Width of Wall Cap on Heel Side (From Abutment Wall Face). 2.525 m

h) Width (Longitudinal Length) of Abutment Well Cap, 5.500 m

i) Length (Transverse Length) of Abutment Well Cap, 12.750 m

j) 10.250 m

RW&D.

PW&B.

hR-Post.

hCurb.

Rnos.

C/CD-R-Post.

tSlab.

tWC

NGirder.

NX-Girder.

hGirder.

hX-Girder.

bGirder.

bX-Girder.

C/CD-Girder.

CD-Ext.-Girder-Edg.

ClD-Int.-Girder.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

ASup-Vert. m2

hWell-Cap.

hSteam.

hb-wall

H-W-Wall

WWell-Cap-Hell

WAb-Cap

LAb-T-W-Cap



Direction.

k) Inner Length of Abutment Wall in between Wing Walls (Transverse), 9.350 m







r) 2.825 m








iii) Dimensions of RCC Well for Foundation.

a) 5.500 m

b) 12.750 m



LAb-T-Inner

t.-Ab-wal-Bot.

t.-Ab-wal-Top.

tWW-Countf.

NW-W-count


SAver-Count&Ab.



t-Wing-wall

tw-wall-Cant.

Lw-wall-Cant.

hw-wall-Cant.-Rec.

hw-wall-Cant.-Tri.

L-W-Cap-Toe.

L-W-Cap-Heel-Aver.




HWell-pro.

tWell-Wall.




h) 4.300 m

i) 7.250 m

j) 4.300 m





o) 2.750 m

p) 63.633

q) Total Length of Staining of Well (Main & Partitions) through Center line 38.494 m

r) 66.879

s) 2.939 m

t) 2.100 m

u) (4.750) m

v) Depth of Well Portion within Bed-rock, 1.500 m

3 Design Data in Respect of Unit Weight & Strength of Materials :

i) Unit Weight of Different Materials :

tWall-Perti

DOuter.




NPock.




Width of Well from its c.g. Line in X-X. = WWell-Y-Y/2 W1/2-Well-Y-Y

Surface Area of Well at Top & Bottom Level = pDOuter2/4 + LRect*DOuter AWell. m2

LStaining.

= p*(DOuter+ DInner)/2 + 2*LRect. + 2*LParti

Surface Area of Well Cap = LAb-T-W-Cap*W1/2-Well-Y-Y + 0.5*pDOuter2/4 AWell-Cap. m2

+ LRect*W1/2-Well-Y-Y


= (LAb-T-W-Cap*(W1/2-Well-Y-Y)2*1.50+ (0.5*pDOuter2/4)*0.50*DOuter*3/4

+ LRect*W1/2-Well-Y-Y2/2)/AWell-Cap.

RL of Height Flood Level (HFL) HFLRL

RL of Maximum Scoring Level (MSL) MSLRL

HB-Rock


i)

9.807






ii)






ii) Design Data for Resistance Factors for Conventional Construction (AASHTO LRFD-5.5.4.2.1). :










j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85


a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa



gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.




c



c Ec






f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa


a) 8.384 n 8

b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102

v) Sub-soil Investigation Report & Bearing Capacity of Bed-rock :

a) N 50 Over

b) 33

c) Recommended Allowable Bearing Capacity of Bed-rock (Soil Investigation p 770

4 Intensity of Different Imposed Loads, Load Coefficients & Multiplier Factors :


a) 0.441

b) f 34



d) 0.34 to 0.45 dim



7.935


c) 0.018





Modular Ratio, n = Es/Ec>6

Value of Ratio of Steel & Concrete Flexural Strength, r = fs/fc = 164/8.400 Value of k = n/(n + r) = 9.000/(9.000 + 20) Value of j = 1 - k/3 = 1 - 0.307/3

Value of R = 0.5*(fckj) = 0.5*(8.400*0.307*0.898) = 1.156



= 15 + 1/2(50 - 15) = 32.5 . Say N/ = 33

kN/m2

Report witht SPT Value 50 over, p = 7.2 Ton/ft2. = 770kN/m2)





O


Dp-ES N/mm2

Dp-ES = ksqs in Mpa. Where; kN/m2







0.004761

4.761



0.004761

4.761

c) k 0.441

d) 1,835.499

e)

f) g 9.807

g) 900.000 mm


h) Width of Live Load Surcharge Pressure for Abutment having 900.000 mm 0.900 m


i) 1,050.000 mm


j) Width of Live Load Surcharge Pressure for Wing Walls, 600.000 mm 0.600 m

600.000 mm 0.600 m


a) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.000800 Mpa

Elements in Lateral Direction of Wind Flow (Parallel to Traffic). 0.800 AASHTO-LRFD-3.8.1.2.2; Table-3.8.1.2.2-1.

b) Horizontal Wind Load Intensity on Vertical Fcaes of Superstructure 0.0009000 Mpa

Elements in Longitudinal Direction of Wind Flow (Perpendicular to Traffic). 0.900


= wE*10-3N/mm2


kN/m2


kN/m2


kN/m2


kN/m2



Since wE, the Unit Wieght of Soil = 18kN/m3, thus gs = wE*10^3/g




Weq-Ab<6.00m.

H < 6000mm.& H ³ 6000mm.

Weq-Ab³6.00m.



Weq-WW<6.00m.

Having H < 6000mm.& H ³ 6000mm.

Weq-WW³6.00m.

pWind-Sup-Let.

kN/m2

pWind-Sup-Long.

kN/m2


a) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.000950 Mpa

0.950

b) Horizontal Wind Load Intensity on Vertical Fcaes of Substructure 0.001645 Mpa

Elements in Longitudinal Direction (Perpendicular to Traffic). 1.645








i) Permanent & Dead Load Multiplier Factors :






pWind-Sub-Let.



pWind-Sub-Long.

kN/m2


pWind-LL-Sup-Let.


pWind-LL-Sup-Long.

action at 1800mm above Deck & Considering 600 Skew Angle of Main Force; for Two Lane Bridge.(AASHTO-LRFD-3.8.1.3; Table- 3.8.1.3-1).

pLL-Sup-Break.

pLL-25%-Truck.

pLL-5%-(Tru+Lane.)



gEH


gEV




ii) Live Load Multiplier Factors :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000



k) 1.000






gES



















q) -

r) -

t) 1.000












b) 1.000


d) 1.000







gEH


gEV


gES






e) 1.000



h) 1.000

i) 1.000



k) 1.000






q) -

r) -

t) 1.000

7 Checkings in Respect of RCC Staining Walls Thickness & Bearing Capacity of Bed-rock :

i) Checking for Thickness of RCC Staining Walls of Well Foundation.

a) Provided Thickness of RCC Staining Walls for Well Foundation. 0.600 m

b)

c) 0.030

















tWell-pro.

Formula for Checking of Wall Thickness, t = KdÖh, Where,

K is Coefficient for Dumb-bell shap Twin-D & One Rectangular Type RCC K


Well Foundation in predominantly Sandy Strata having value = 0.030.

d) d 5.500 m

e) h 6.325 m

f) 0.415 m

g)

ii) Checking Bearing Capacity of Bed-rock underneath RCC Well Foundation against all Imposed VerticalLoads from Superstructure & Substructure (DL & LL) :

a) Dead Load Reaction from Total Super-Structure 2,185.716 kN

b) 1,108.714 kN

c) Vertical Loads from Sub-structural Componts, Soil, Surcharge etc. 5,729.857 kN

d) Total Vertical Forces due to Dead & Live Load 9,024.287 kN

e) Bearing Capacity of Bea-rock as per Soil Investigation Report 770.000

f) 141.817

g) Relation between Bearing Capacity of Beadrock & Imposed Pressure through Well Satisfy OK

h)Bridge Structure.

8 Computation of Loads (DL & LL) from Superstructure, Substructure,Soil & Surcharge Elements :

i) Vertical Loads on RCC Well from Superstructure & Substructure Components, Soil & Surcharge :

a)the Staining of RCC Well, thus it is considered an uniform intensity of Vertical Loads upon the Walls with equal magnitude per meter Length of Staining.

b) 1,976.278 kN

c) 209.438 kN

d) 3,627.592 kN

d = DOuter the Outer Diameter which is the Smaller dimension of RCC Well

h = HWell-pro, Depth of Well (Since GL > LWL)

Thus the Required thickness of Well Wall, t = KdÖh treq.

Since the Provided thickness for RCC Staining Walls of Well tWall > treq. thus the Provision is O.K.

RDL-Supr.

All Live Load Reaction (Wheel+IM Load, Lane Load & Pedestrian Loads) RLL-Super.

RLL-Sub&Soil.

PV

pBed-rock kN/m2

Per m2 Load on Bed-rock from Superstructure & Substructure (DL & LL) pImpos. kN/m2

through RCC Well for Bridge Foundation = Total Load/ X- Area of Well

= PV/AWell

Since Bed-rock Bearing Capacity pBed-rock > pImpos, thus it is capable to Uphold the Imposed Loads of

Since all the Vertical Loads (DL & LL) from Superstructure & Substructure will transmit to the Bed-rock through

Vertical DL Reaction on Well from Superstructure without Utility & WC RDL-V-Supr.

Vertical DL Reaction on Well from Utility & WC on Superstructure RDL-V-Ut.&WC.

Vertical DL Reaction on Well from Substructure Components. RDL-V-Sub.


e) 2,102.265 kN

f) 574.221 kN

g) 232.500 kN

h) 112.500 kN

i) 247.398 kN

j) 16.139 kN

ii) Horizontal Loads (DL & LL)on Superstructure & Substructure due to Lateral Siol Pressure & others:

a) 995.173 kNFaces of Abutment Wall & Wing Walls (Parallel to Traffic).

b) 600.557 kNWell Cap for Rectangular Pressure Block (Parallel to Traffic).

c) 72.839 kNFace of Well Cap for Triangular Pressure Block (Perpendicular to Traffic).

d) 402.334 kNAbutmen Wall & Wing Walls (Perpendicular to Traffic).

e) 121.398 kNWell Cap (Perpendicular to Traffic).

f) 362.101 kNAbutmen Wall & Wing Walls (Perpendicular to Traffic).

g) 109.258 kNWall Cap (Perpendicular to Traffic).

h) 13.750 kN

Vertical DL Reaction on Well from Soil Load upon Well Cap. RDL-V-Soil.

Vertical LL Reaction on Well due to Wheel Load (Without IM) RLL-V- Wheel.

Vertical LL Reaction on Well due to Lane Load on Superstructure RLL-V-Lane.

Vertical LL Reaction on Well due to Pedestrian Load RLL-V-Pedes.

Vertical Reaction due to Surcharge LL on Well Cap RLL-V-Sur-Ab.

on Abutment Wall Face =Dp-LL-Ab*Weq-Ab*LAb-T-Inner

Vertical Reaction due to Surcharge LL on Well Cap RLL-V-Sur-WW.

on Wing Wall Face =2*Dp-LL-WW*Weq-WW*LW-Cap-Heel-Aver.

Horizontal Dead Load (DL) due to Lateral Soil Pressure on Vertical RDL-H-Soil-Ab

= 0.5*k0*wE*H12*LAB-Trans.

Horizontal Surcharge Dead Load (DL) Pressure on Vertical Face of RDL-H-Soil-Cap-Rec.

= k0*wE*H1*hWell-Cap*LAb-T-W-Cap.

Horizontal Dead Load (DL) due to Lateral Soil Pressure on Vertical RDL-H-Soil-Cap-Tri.

=0.5* ko*gE*(hWell-Cap)2*LAb-T-W-Cap.

Horizontal Dead Load (DL) Surcharge Pressure on Vertical Faces of RDL-H-Sur-Ab

= Dp-ES*H1*LAB-Trans.

HorizontalDead Load (DL) Surcharge Pressure on Vertical Face of RDL-H-Sur-W-Cap

= Dp-ES*hWell-Cap*LAb-T-W-Cap.

Horizontal Surcharge Live Load (LL) Pressure on Vertical Faces of RLL-H-Sur-Ab

= Dp-LL-Ab*H1*LAB-Trans.

Horizontal Surcharge Live Load (LL) Pressure on Vertical Face of RLL-H-Sur-Ab

= Dp-LL-Ab*hWell-Cap*LAb-T-W-Cap.

Horizontal Live Load (LL) due to Wind on Superstructure RLL-H-Wind-Super.


i) 48.171 kN

j) 13.750 kN

k) 162.500 kN

9 Computation of Factored Loads (DL & LL) under Strength Limit State Design (USD) from Superstructure, Substructure, Soil & Surcharge Elements :

i) Factored Vertical Loads on RCC Well :

a) 2,732.145 kN

b) 314.156 kN

c) 4,534.490 kN

d) 2,838.057 kN

e) 1,004.887 kN

f) 406.875 kN

g) 196.875 kN

h) 432.947 kN

i) 28.243 kN

k) Total Factored Vertical Loads upon RCC Well. 12,488.676 kN

= pWind-Sup-Let.*ASup-Vert.

Horizontal Live Load (LL) due to Wind on Substructure RLL-H-Wind-Sub.

= pWind-Sub-Let.*H1*LAb-Trans.

Horizontal Live Load (LL) due to Wind on Live Load RLL-H-Wind-LL.

= pWind-LL-Sup-Let.*LGir.

Horizontal Live Load (LL) due to Breaking Force of Vehicle on RLL-H-Break.

Bridge Deck =pLL-Sup-Break.

Factored Vertical DL from Superstructure without Utility & WC FRDL-V-Supr.-USD

= gDC*RDL-V-Supr.

Factored Vertical DL due to Utility & WC FRDL-V-Ut.&WC.-USD

= gDW*RDL-V-Ut.&WC.

Factored Vertical DL from Substructure Components. FRDL--V-Sub.-USD

= gDC*RDL--V-Sub.

Factored Vertical DL from Soil Load upon Well Cap. FRDL--V-Soil-USD.

= gEV*RDL--V-Soil.

Factored Vertical LL on Well due to Wheel Load (Including IM) FRLL-V- Wheel.USD

= m*gLL-Truck*RLL-V- Wheel.

Factored Vertical LL on Well due to Lane Load on Superstructure FRLL-V-Lane.-USD

= m*gLL-Lane*RLL-V-Lane.

Factored Vertical LL on Well due to Pedestrian Load FRLL-V-Pedes.-USD

= m*gLL-Pedes*RLL-V-Pedes.

Factored Vertical LL due to Surcharge on Abutment Wall Face FRLL-V-Sur-Ab.-USD

= m*gLL-Sur.*RLL-V-Sur-Ab.

Factored Vertical LL due to Surcharge on Wing Wall Face FRLL-V-Sur-WW.-USD

= m*gLL-Sur.*RLL-V-Sur-WW.

åFV-Load.-USD


l) Factored Vertical Load on per meter Length of Staining of Well (Including 324.433 kN/m

ii) Factored Horizontal Loads on Superstructure & Substructure:

a) 1,492.760 kNWall & Wing Walls (Parallel to Traffic).

b) 900.836 kNFace of Well Cap for Rectangular Pressure Block (Parallel to Traffic).

c) 109.258 kNFace of Well Cap for Triangular Pressure Block (Perpendicular to Traffic).

d) 603.501 kNAbutmen Wall & Wing Walls (Perpendicular to Traffic).

e) 72.257 kNWell Cap (Perpendicular to Traffic).

f) 633.676 kNAbutmen Wall & Wing Walls (Perpendicular to Traffic).

g) 191.202 kNWall Cap (Perpendicular to Traffic).

h) 19.250 kN

i) 67.440 kN

j) 13.750 kN

k) 284.375 kN

pStaining.USD

Partition Walls) = åFV-Load./LStaining.

Factored Horizontal DL due to Lateral Soil Pressure on Abutment FRDL-H-Soil-Ab-USD

=gEH*RDL-H-Soil-Ab.

Factored Horizontal DL due to Lateral Soil Pressure on Vertical FRDL-H-Soil-Cap-ReC.-USD

= gEH*RDL-H-Soil-Cap-Rec.

Factored Horizontal DL due to Lateral Soil Pressure on Vertical FRDL-H-Soil-Cap-Tri.-USD

= gEH*RDL-H-Soil-Cap-Tri.

Factored Horizontal DL Surcharge Pressure on Vertical Faces of FRDL-H-Sur-Ab-USD

= gES*RDL-H-Sur-Ab

Factored Horizontal DL Surcharge Pressure on Vertical Face of FRDL-H-Sur-W-Cap-USD

= gES*RDL-H-Sur-W-Cap

Factored Horizontal Surcharge LL Pressure on Vertical Faces of FRLL-H-Sur-Ab-USD

= gLL-LS.*RLL-H-Sur-Ab

Factored Horizontal Surcharge LL Pressure on Vertical Face of FRLL-H-Sur-W-Cap-USD

= gLL-LS.*RLL-H-Sur-W-Cap

Factored Horizontal LL due to Wind on Superstructure FRLL-H-Wind-Super.-USD

= gLL-WS.*RLL-H-Wind-Super.

Factored Horizontal LL due to Wind on Substructure FRLL-H-Wind-Sub.-USD

= gLL-WS.*RLL-H-Wind-Sub.

Factored Horizontal LL due to Wind on Live Load FRLL-H-Wind-LL.-USD

= m*gLL-WL.*RLL-H-Wind-LL.

Factored Horizontal LL due to Breaking Force of Vehicle on Bridge FRLL-H-Break.-USD

Deck = m*gLL-BR.*RLL-H-Break.


l) Total Factored Horizontal Pressure 4,388.306 kN

m) 344.181 kN/m

10 Horizontal Loads (DL & LL) on Faces of RCC Well for per meter Length at Top & Bedrock Level:

i) Intensity Horizontal Earth Loads on Vertical Face at Top Level of Well (Perpendicular to Traffic) :

a) 48.774 kN/m

b) 7.935 kN/m

c) 4.761 kN/m

d) 56.708 kN/m

ii) Intensity Horizontal Earth Loads on Vertical Face at Bedrock Level of Well (Perpendicular to Traffic) :

a) 87.058 kN/m

b) 7.935 kN/m

c) 4.761 kN/m

c) 94.992 kN/m

11 Factored Horizontal Loads (DL & LL) on Faces of RCC Well under Strength Limit State Design (USD) :

i) Factored Horizontal Load Intensity due to Soil & Surcharge at Top Level :

a) 73.160 kN/m

b) 11.902 kN/m

c) 8.331 kN/m

SPH-Well-Cap-USD

Factored Horizontal Pressure per meter Legnth = SPH-Well-Cap-USD/LWEll-X-X pH-Y-Y-USD

Horizontal Earth Load (DL) Itensity on Earth Face at Well's Top Level PH-Soil-Top.

(Just below Well Cap) = ko*wE*H kN/m

Horizontal Dead Load Surcharge (DL) Itensity on Earth Face at PH-Sur-DL-Top.

Well's Top Level (Just below Well Cap) = Dp-ES kN/m

Horizontal Live Load Surcharge (LL) Itensity on Earth Face at PH-Sur-LL-Top.

Well's Top Level (Just below Well Cap) = Dp-LL-Ab³6.00m kN/m

Intensity of Total Horizontal Load (DL+ LL) at Well's Top Level. åPH-Top-UF

Horizontal Earth Load (DL) Itensity on Earth Face at Bed-rock PH-Soil-Rock

Level = ko*wE*(H + HWell-pro.- H-B-Rock) kN/m

Horizontal Dead Load Surcharge (DL) Itensity on Earth Face at PH-Sur-DL-Rock

Bed-rock Level = Dp-ES kN/m

Horizontal Live Load Surcharge (LL) Itensity on Earth Face at PH-Sur-LL-Rock.

Bed-rock Level = Dp-LL-Ab³6.00m kN/m

Intensity of Total Horizontal Load (DL + LL) at Well's Bedrock Level. åPH-Rock-UF

Factored Itensity of DL Earth Pressure at Well's Top Level FPH-Soil-Top.-USD

= gEH*PH-Soill-Top. kN/m

Factored Itensity of DL Surcharge Pressure at Well's Top Level. FPH-Sur-DL-Top.-USD

= gES*PH-Sur-DL-Top. kN/m

Factored Itensity of LL Surcharge Pressure at Well's Top Level. FPH-Sur-LL-Top.-USD

= mgLL-LS*PH-Sur-LL-Top. kN/m


d) 85.062 kN/m

ii) Factored Horizontal Load Intensity due to Soil & Surcharge at Bedrock Level :

a) 130.586 kN/m

b) 11.902 kN/m

c) 8.331 kN/m

c) 142.488 kN/m

12 Computation of Coefficients for Calculation of Moments on Well Wall :

i) Span Ratio of Well Wall in respect of of Horizontal to Vertical Spans :

a) Horizontal Span Length of Well Wall (Inner Distance between Partations) 3.450 m

b) Vertical Height of Well Wall (From Bedrock Top upto Well Cap Bottom) 4.825 m

c) Bach Wall Span Ratio for of Horizontal to Vertical with Fixed on Top 0.715

d) Bach Wall Span Ratio for of Vertical to Horizontal with Fixed on Top 1.399

ii) Coefficients for Calculation of Moments related to Triangular Loadings on Well Wall :(Using Table-53 of Reinforced Concrete Design Handbook UK.)

a) k 0.715

b) (+) ve Moment Coefficient of Well Wall for Vertical Span at Top Level. 0.010

c) (+) ve Moment Coefficient of Well Wall for Vertical Span at Bottom. 0.005

d) (+) ve Moment Coefficient of Well Wall for Horizontal Span Strip at Middle. 0.020

e) (-) ve Moment Coefficient of Well Wall for Vertical Span Strip at Bottom 0.010

f) (-) ve Moment Coefficient of Well Wall for Horizontal Span Strip on Faces 0.025

Intensity of Factored Total Horizontal Load (DD + LL) at Well's Top. åFPH-Top.-USD

Factored Itensity of DL Earth Pressure at Well's Bedrock Level FPH-Soil-Rock-USD

= gEH*PH-Soil-Rock. kN/m

Factored Itensity of DL Surcharge Pressure at Well's Bedrock FPH-Sur-DL-Rock.-USD

Level = gES*PH-Sur-DL-Rock. kN/m

Factored Itensity of LL Surcharge Pressure at Well's Bedrock FPH-Sur-LL-Rock.-USD

Level = mgLL-LS*PH-Sur-LL-Rock. kN/m

Intensity of Factored Total Horizontal Load (DD & LL) at Bed-rock. åFPH-Rock.-USD

SL-Well.

= SPocket-X-X-Central.

hRock-Well

=HWell-pro. - HB-Rock

kFixed.-H/V

= SL-Well/hB-Rock-Well

kFixed-V/H

= hB-Rock-Well/SL-Well

kFixed.-H/V = 0.715

(+)cS-V-T-Top

(+)cS-V-T-Bot

(+)cS-H-T

(-)cS-V-T

(-)cS-H-T


iii) Coefficients for Calculation of Moments related to Uniformly Loadings on Well Wall :(Using Table-4.2; 4.3 & 4.4 of Book Design of Concrete Structures- G.Winter)

a) k 1.000





f) 0.045 Well Wall Vertical Span Strip

g) 0.045 Back Wall Horizontal Span Strip

iv) Coefficients for Calculation of Shearing Froces on Well Wall Vertical Faces :(Using Table-4.5 of Book Design of Concrete Structures- G.Winter)

a) k 1.000

b) Coefficiant of Shearing Forces for Horizontal Span Strip having Shearing 0.500 Forces on Well Wall Vertical Faces

c) Coefficiant of Shearing Forces for Vertical Span Strip having Shearing 0.500 Forces on Well Wall Horizontal Face at Bedrock

13 Factored Horizontal Moments on Face of Well Wall under Strength Limit State of Design (USD):

i) Philosophy in Calculation of Moment & Shear :

a) Moments on Horizontal Span of Well due to Imposed Horizontal Loads are to be Calculate Considering 1(One)meter wide strips of Well Wall in Vertical & Fixed Ends in both Horizontal & Vertical Faces with Well Partitions &Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition Walls. The.Moments & Shearing Forces will be based upon Respective Coefficients.

b) 3.450 m

ii)

Since kFixed-V/H = 1.399 , thus let Consider kFixed-V/H = 1.000 (Case-2)

(+) ve Moment Coefficient for Dead Load (DL) on Well Wall Vertical (+)cS-V-U-DL

(+) ve Moment Coefficient for Dead Load (DL) on Well Wall Horizontal (+)cS-H-U-DL

(+) ve Moment Coefficient for Live Load (LL) on Well Wall Vertical (+)cS-V-U-LL

(+) ve Moment Coefficient for LiveLoad (LL) on Well Wall Horizontal (+)cS-H-U-LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-V-U-DL+LL

(-) ve Moment Coefficient for Dead Load & Live Load (DL + LL) on (-)cS-H-U-DL+LL

Since kFixed-V/H = 1.399 , thus let Consider kFixed-V/H = 1.000 (Case-2)

cShear-H-DL+LL

cShear-V-DL+LL

Horizontal Span Length = SPocket-X-X-Central. SL-Well

(+) Moments on Mid of Well Wall Horizontal Span at Top Level due to Applied Lateral Loads :


a) (+) ve Moment due to Lateral Soil Pressure upon Well Wall 17.416 kN-m/m

17.416*10^6 N-mm/m

b) 2.550 kN-m/m

2.550*10^6 N-mm/m

c) 2.677 kN-m/m

2.677*10^6 N-mm/m

d) Total Horizontal (+) ve Moment atWell Top Level 22.643 kN-m/m 22.643*10^6 N-mm/m

iii)


31.086*10^6 N-mm/m

b) 2.550 kN-m/m

2.550*10^6 N-mm/m

c) 2.677 kN-m/m

2.677*10^6 N-mm/m

d) Total Horizontal (+) ve Moment at Bedrock Level 36.313 kN-m/m

36.313*10^6 N-mm/m

iv)

a) (-) ve Moment due to Lateral Soil Pressure upon Well Wall 21.770 kN-m/m

21.770*10^6 N-mm/m

b) 6.375 kN-m/m

6.375*10^6 N-mm/m

c) 4.462 kN-m/m

2.677*10^6 N-mm/m

d) Total Horizontal (-) ve Moment atWell Top Level 32.607 kN-m/m

22.643*10^6 N-mm/m

v)


(+) MH-Soil-Top.-USD

= (+)cS-H-T*FPH-Soil-Top.-USD*SL-Well.2

(+) ve Moment due to Lateral Dead Load (DL) Surcharge (+) MH-Sur-DL-Top.-USD

Pressure upon Well Wall = (+)cS-H-U-DL*FPH-Sur-DL-Top.-USD*SL-Well.2

(+) ve Moment due to Lateral Live Load (LL) Surcharge (+) MH-Sur-LL-Top.-USD

Wall = (+)cS-H-U-LL*FPH-Sur-LL-Top.-USD*SL-Well.2

(+) MTotal-H-Top-USD

(+) Moments on Mid of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Loads :

(+) MH-Soil-Rock-USD

= (+)cS-H-T*FPH-Soil-Rock-USD*SL-Well.2

(+) ve Moment due to Dead Load (DL) Surcharge Pressure upon (+) MH-Sur-DL-Rock-USD

Well Wall = (+)cS-H-U-DL*FPH-DL-Sur-Rock.-USD*SL-Well.2

(+) ve Moment due to Live Load (LL) Surcharge Pressure upon (+) MH-Sur-LL-Rock-USD

Wall = (+)cS-H-U-LL*FPH-LL-Sur-Rock.-USD*SL-Well.2

(+) MTotal-H-Rock-USD

(-) Moments on Faces of Well Wall Horizontal Span at Top Level due to Applied Lateral Loads :

(-) MH-Soill-Top.-USD

= (-)cS-H-T*FPH-Soil-Top.-USD*SL-Well.2

(-) ve Moment due to Lateral Dead Load (DL) Surcharge (-) MH-Sur-DL-Top.-USD

Pressure upon Well Wall = (-)cS-H-U-DL+LL*FPH-Sur-DL-Top.-USD*SL-Well.2

(-) ve Moment due to Lateral Live Load (LL) Surcharge (-) MH-Sur-LL-Top.-USD

Wall = (-)cS-H-U-DL+LL*FPH-Sur-LL-Top.-USD*SL-Well.2

(-) MTotal-H-Top-USD

(-) Moments on Faces of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Loads :

(-) MH-Soil-Rock-USD


38.858*10^6 N-mm/m

b) 6.375 kN-m/m

6.375*10^6 N-mm/m

c) 4.462 kN-m/m

4.462*10^6 N-mm/m

d) Total Horizontal (-) ve Moment at Bedrock Level 49.695 kN-m/m

49.695*10^6 N-mm/m

14 Factored Vertical Moments on Bedrock Face of Well Wall under Strength Limit State of Design (USD):


a) Moments on Vertical Span of Well due to Imposed Horizontal Loads at Bedrock are to be Calculate Considering

1(One) meter wide strips of Well Wall in Horizontal & Fixed Ends in both Horizontal & Vertical Faces with Well

Partitions & Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition

Walls. The Moments & Shearing Forces will be based upon Respective Coefficients.

b) 4.825 m

ii)


15.201*10^6 N-mm/m

b) 4.987 kN-m/m

4.987*10^6 N-mm/m

c) 3.491 kN-m/m

3.491*10^6 N-mm/m

d) Total Horizontal (+) ve Moment atWell Top Level 23.679 kN-m/m

23.679*10^6 N-mm/m

iii)


20.268*10^6 N-mm/m

b) 12.469 kN-m/m

12.469*10^6 N-mm/m

= (-)cS-H-T*FPH-Soil-Rock-USD*SL-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MH-Sur-DL-Rock-USD

Well Wall = (-)cS-H-U-DL+LL*FPH-DL-Sur-Rock.-USD*SL-Well.2

(-) ve Moment due to Live Load (LL) Surcharge Pressure upon (-) MH-Sur-LL-Rock-USD

Wall = (-)cS-H-U-LL+DL*FPH-LL-Sur-Rock.-USD*SL-Well.2

(-) MTotal-H-Rock-USD

Vertical Span Length = hRock-Well hRock-Well

(+) Moments on Mid of Well Wall Vertical Span due to Applied Lateral Loads :

(+) MV-Soil-USD

= (+)cS-V-T*FPH-Soil-Rock.-USD*hRock-Well.2

(+) ve Moment due to Lateral Dead Load (DL) Surcharge (+) MV-Sur-DL-USD

Pressure upon Well Wall = (+)cS-V-U-DL*FPH-Sur-DL-Rock.-USD*hRock-Well.2

(+) ve Moment due to Lateral Live Load (LL) Surcharge (+) MV-Sur-LL-USD

Wall = (+)cS-V-U-LL*FPH-Sur-LL-Rock.-USD*hRock-Well.2

(+) MTotal-V-Rock-USD

(-) Moments on Faces of Well Wall Vertical Span at Bedrock Level due to Applied Lateral Loads :

(-) MV-Soil-Rock-USD

= (-)cS-V-T*FPH-Soil-Rock-USD*hRock-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MV-Sur-DL-Rock-USD

Well Wall = (-)cS-V-U-DL*FPH-DL-Sur-Rock.-USD*hRock-Well.2


c) 8.728 kN-m/m

8.728*10^6 N-mm/m


41.464*10^6 N-mm/m

15 Factored Shearing Forces on both Horizontal & Vertical Span of Well Wall Support Faces under Strength

Limit State of Design (USD):

i) Shearing Forces in Horizontal Span at Bottom Level on Faces of Partition Walls :

a) The Maximum Shearing Forces occoures at Bottom Level on Partion 122.896 kN/m

122.896*10^3 N/m

ii) Shearing Forces in Vertical Span at Bottom Level on Bedrock Level :

a) The Maximum Shearing Forces occoures at Bottom Level on Bedrock 171.876 kN/m

171.876*10^3 N/m

16 Factored Horizontal Loads (DL & LL) on Faces of RCC Well under Strength Limit State Design (WSD) :

i) Factored Horizontal Load Intensity due to Soil & Surcharge at Top Level :

a) 48.774 kN/m

b) 7.935 kN/m

c) 4.761 kN/m

d) 61.469 kN/m

ii) Factored Horizontal Load Intensity due to Soil & Surcharge at Bed-rock Level :

a) 87.058 kN/m

b) 7.935 kN/m

c) 4.761 kN/m

c) 99.753 kN/m

(-) ve Moment due to Live Load (LL) Surcharge Pressure upon (-) MV-Sur-LL-Rock-USD

Wall = (-)cS-H-U-LL+DL*FPH-LL-Sur-Rock.-USD*hRock-Well.2

(-) MTotal-V-Rock-USD

VH-Wall-Rock-USD

Wall Faces having value = cShear-H-DL+LL*åFPH-B-Rock.-USD*SL-Well

VV-Rock-USD

Face having value = cShear-VDL+LL*åFPH-B-Rock.-USD*hRock-Well

Factored Itensity of DL Earth Pressure at Well's Top Level FPH-Soil-Top.-WSD

= gEH*PH-Soil-Well-Top. kN/m

Factored Itensity of DL Surcharge Pressure at Well's Top Level. FPH-DL-Sur-Top.-WSD

= gES*PH-DL-Sur-Well-Top. kN/m

Factored Itensity of LL Surcharge Pressure at Well's Top Level. FPH-Sur-LL-Top.-WSD

= mgLL-LS*PH-Sur-LL-Top. kN/m

Intensity of Factored Total Horizontal Load (DD & LL) at Well's Top. åFPH-Top.-WSD

Factored Itensity of DL Earth Pressure at Well's Bottom Level FPH-Soil-Rock-WSD

= gEH*PH-Soil-Rock. kN/m

Factored Itensity of DL Surcharge Pressure at Well's Bottom FPH-Sur-DL-Rock.-WSD

Level = gES*PH-DL-Sur-Rock. kN/m

Factored Itensity of LL Surcharge Pressure at Well's Bedrock FPH-Sur-LL-Rock.-WSD

Level = mgLL-LS*PH-Sur-LL-Rock. kN/m

Intensity of Factored Total Horizontal Load (DD & LL) at Bed-rock. åFPH-Rock.-WSD


17 Factored Horizontal Moments on Outer Face of Well Wall under Service Limit State of Design (WSD):


a) Moments on Horizontal Span of Well due to Imposed Horizontal Loads are to be Calculate Considering 1(One)meter wide strips of Well Wall in Vertical & Fixed Ends in both Horizontal & Vertical Faces with Well Partitions &Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition Walls. The.

Moments & Shearing Forces will be based upon Respective Coefficients.

b) 3.450 m

ii)


11.611*10^6 N-mm/m

b) 1.700 kN-m/m

1.700*10^6 N-mm/m

c) 1.530 kN-m/m

1.530*10^6 N-mm/m


14.840*10^6 N-mm/m

iii)


20.724*10^6 N-mm/m

b) 1.700 kN-m/m

1.700*10^6 N-mm/m

c) 1.530 kN-m/m

1530*10^6 N-mm/m

d) Total Horizontal (+) ve Moment at Bedrock Level 23.954 kN-m/m

23.954*10^6 N-mm/m

iv)


14.513*10^6 N-mm/m

b) 4.250 kN-m/m


(+) Moments on Mid of Well Wall Horizontal Span at Top Level due to Applied Lateral Loads :

(+) MH-Soil-Top.-wSD

= (+)cS-H-T*FPH-Soil-Top.-wSD*SL-Well.2

(+) ve Moment due to Lateral Dead Load (DL) Surcharge (+) MH-Sur-DL-Top.-WSD

Pressure upon Well Wall = (+)cS-H-U-DL*FPH-Sur-DL-Top.-WSD*SL-Well.2

(+) ve Moment due to Lateral Live Load (LL) Surcharge (+) MH-Sur-LL-Top.-WSD

Wall = (+)cS-H-U-LL*FPH-Sur-LL-Top.-WSD*SL-Well.2

(+) MTotal-H-Top-WSD

(+) Moments on Mid of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Loads :

(+) MH-Soil-Rock-WSD

= (+)cS-H-T*FPH-Soil-Rock-WSD*SL-Well.2

(+) ve Moment due to Dead Load (DL) Surcharge Pressure upon (+) MH-Sur-DL-Rock-WSD

Well Wall = (+)cS-H-U-DL*FPH-DL-Sur-Rock.-WSD*SL-Well.2

(+) ve Moment due to Live Load (LL) Surcharge Pressure upon (+) MH-Sur-LL-Rock-WSD

Wall = (+)cS-H-U-LL*FPH-LL-Sur-Rock.-WSD*SL-Well.2

(+) MTotal-H-Rock-WSD

(-) Moments on Faces of Well Wall Horizontal Span at Top Level due to Applied Lateral Loads :

(-) MH-Soill-Top.-WSD

= (-)cS-H-T*FPH-Soil-Top.-wSD*SL-Well.2

(-) ve Moment due to Lateral Dead Load (DL) Surcharge (-) MH-Sur-DL-Top.-WSD


4.250*10^6 N-mm/m

c) 2.550 kN-m/m

2.550*10^6 N-mm/m

d) Total Horizontal (-) ve Moment atWell Top Level 21.313 kN-m/m

21.313*10^6 N-mm/m

v)


25.905*10^6 N-mm/m

b) 4.250 kN-m/m

4.250*10^6 N-mm/m

c) 2.550 kN-m/m

2.550*10^6 N-mm/m


32.705*10^6 N-mm/m

18 Factored Vertical Moments on Bedrock Face of Well Wall under Service Limit State of Design (WSD):


a) Moments on Vertical Span of Well due to Imposed Horizontal Loads at Bedrock are to be Calculate Considering 1(One) meter wide strips of Well Wall in Horizontal & Fixed Ends in both Horizontal & Vertical Faces with Well Partitions & Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition Walls. The Moments & Shearing Forces will be based upon Respective Coefficients.

b) 4.825 m

ii)


10.134*10^6 N-mm/m

b) 3.325 kN-m/m

3.325*10^6 N-mm/m

c) 2.992 kN-m/m

2.992*10^6 N-mm/m


16.451*10^6 N-mm/m

Pressure upon Well Wall = (-)cS-H-U-DL+LL*FPH-Sur-DL-Top.-WSD*SL-Well.2

(-) ve Moment due to Lateral Live Load (LL) Surcharge (-) MH-Sur-LL-Top.-WSD

Wall = (-)cS-H-U-DL+LL*FPH-Sur-LL-Top.-WSD*SL-Well.2

(-) MTotal-H-Top-WSD

(-) Moments on Faces of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Loads :

(-) MH-Soil-Rock-WSD

= (-)cS-H-T*FPH-Soil-Rock-WSD*SL-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MH-Sur-DL-Rock-WSD

Well Wall = (-)cS-H-U-DL+LL*FPH-DL-Sur-Rock.-WSD*SL-Well.2

(-) ve Moment due to Live Load (LL) Surcharge Pressure upon (-) MH-Sur-LL-Rock-WSD

Wall = (-)cS-H-U-LL+DL*FPH-LL-Sur-Rock.-WSD*SL-Well.2

(-) MTotal-H-Rock-WSD


(+) Moments on Mid of Well Wall Vertical Span due to Applied Lateral Loads :

(+) MV-Soil-WSD

= (+)cS-V-T-Bot*FPH-Soil-Rock.-WSD*hRock-Well.2

(+) ve Moment due to Lateral Dead Load (DL) Surcharge (+) MV-Sur-DL-WSD

Pressure upon Well Wall = (+)cS-V-U-DL*FPH-Sur-DL-Rock.-WSD*hRock-Well.2

(+) ve Moment due to Lateral Live Load (LL) Surcharge (+) MV-Sur-LL-WSD

Wall = (+)cS-V-U-LL*FPH-Sur-LL-Rock.-WSD*hRock-Well.2



iii)


20.268*10^6 N-mm/m

b) 8.312 kN-m/m

8.312*10^6 N-mm/m

c) 4.987 kN-m/m

4.987*10^6 N-mm/m

d) Total Horizontal (-) ve Moment at Bedrock Level 33.567 kN-m/m 33.567*10^6 N-mm/m

19 Unfactored Dead Load (DL) Horizontal Moments on Bedrock Face of Well Wall :


a) Moments on Horizontal Span of Well due to Imposed Horizontal Loads are to be Calculate Considering 1(One)meter wide strips of Well Wall in Vertical & Fixed Ends in both Horizontal & Vertical Faces with Well Partitions &Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition Walls. The.Moments & Shearing Forces will be based upon Respective Coefficients.

b) 3.450 m

ii)


20.724*10^6 N-mm/m

b) 1.700 kN-m/m

1.700*10^6 N-mm/m

c) Total Horizontal (+) ve Moment at Bedrock Level 22.424 kN-m/m

22.424*10^6 N-mm/m

iii)


25.905*10^6 N-mm/m

b) 4.250 kN-m/m

4.250*10^6 N-mm/m

c) Total Horizontal (-) ve Moment at Bedrock Level 30.155 kN-m/m

30.155*10^6 N-mm/m

(-) Moments on Faces of Well Wall Vertical Span at Bedrock Level due to Applied Lateral Loads :

(-) MV-Soil-Rock-WSD

= (-)cS-V-T*FPH-Soil-Rock-WSD*hRock-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MV-Sur-DL-Rock-WSD

Well Wall = (-)cS-V-U-DL+LL*FPH-DL-Sur-Rock.-WSD*hRock-Well.2

(-) ve Moment due to Live Load (LL) Surcharge Pressure upon (-) MV-Sur-LL-Rock-WSD

Wall = (-)cS-H-U-LL+DL*FPH-LL-Sur-Rock.-WSD*hRock-Well.2

(-) MTotal-V-Rock-WSD


(+) Moments on Mid of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Dead Loads :

(+) MH-Soil-Rock-UF

= (+)cS-H-T*PH-Soil-Rock*SL-Well.2

(+) ve Moment due to Dead Load (DL) Surcharge Pressure upon (+) MH-Sur-DL-Rock-UF

Well Wall = (+)cS-H-U-DL*PH-DL-Sur-Rock.*SL-Well.2

(+) MTotal-H-Rock-UF

(-) Moments on Faces of Well Wall Horizontal Span at Bedrock Level due to Applied Lateral Dead Loads :

(-) MH-Soil-Rock-UF

= (-)cS-H-T*PH-Soil-Rock*SL-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MH-Sur-DL-Rock-UF

Well Wall = (-)cS-H-U-DL+LL*PH-DL-Sur-Rock.*SL-Well.2

(-) MTotal-H-Rock-UF


20 Unfactored Dead Load (DL) Vertical Moments on Bedrock Face of Well Wall :


a) Moments on Vertical Span of Well due to Imposed Horizontal Loads at Bedrock are to be Calculate Considering 1(One) meter wide strips of Well Wall in Horizontal & Fixed Ends in both Horizontal & Vertical Faces with Well Partitions & Bedrock at Bottom & Well Cap on Top. The Horizontal Span Length is the C/C distance of Partition Walls. The Moments & Shearing Forces will be based upon Respective Coefficients.

b) 4.825 m

ii)


10.134*10^6 N-mm/m

b) 3.325 kN-m/m

3.325*10^6 N-mm/m

c) Total Horizontal (+) ve Moment atWell Top Level 13.459 kN-m/m 13.459*10^6 N-mm/m

iii)


20.268*10^6 N-mm/m

b) 8.312 kN-m/m

8.312*10^6 N-mm/m

c) Total Horizontal (-) ve Moment at Bedrock Level 28.580 kN-m/m 28.580*10^6 N-mm/m

21 Events related to Flexural Design of Horizontal & Vertical Reinforcements of Well Walls both Inner & Outer Faces :

i) Events for Provision of Reinforcements on RCC Well Walls :

a) Thicness of Well Wall, 600.000 mm 0.600 m

b) 18 mm

c) 18 mm

d) 14 mm


(+) Moments on Mid of Well Wall Vertical Span due to Applied Lateral Dead Loads :

(+) MV-Soil-UF

= (+)cS-V-T-Bot*PH-Soil-Rock.*hRock-Well.2

(+) ve Moment due to Lateral Dead Load (DL) Surcharge (+) MV-Sur-DL-UF

Pressure upon Well Wall = (+)cS-V-U-DL*PH-Sur-DL-Rock.*hRock-Well.2


(-) Moments on Faces of Well Wall Vertical Span at Bedrock Level due to Applied Lateral Dead Loads :

(-) MV-Soil-Rock-UF

= (-)cS-V-T*FPH-Soil-Rock*hRock-Well.2

(-) ve Moment due to Dead Load (DL) Surcharge Pressure upon (-) MV-Sur-DL-Rock-UF

Well Wall = (-)cS-V-U-DL+LL*FPH-DL-Sur-Rock.*hRock-Well.2

(-) MTotal-V-Rock-UF

tWall

Let Provide 18f bars as Horizontal Reinforcements on Inner Face of Well. DBar-Hor.-Inner

Let Provide 18f bars as Horizontal Reinforcements on Outer Face of Well. DBar-Hor.-Outer

Let Provide 14f bars as Vertical Reinforcements on Inner Face of Well. DBar-Ver.-Inner


e) 14 mm

f) Let Provide 50mm Clear Cover of Well Walls on Inner Face, 50 mm

g) Let Provide 50mm Clear Cover of Well Walls on Outer Face, 75 mm

h) Effective Depth for Horizontal Tensial Reinforcement on Inner Faces of Wall 559.000 mm

0.559 m

i) Effective Depth for Horizontal Tensial Reinforcement on Outer Faces of Wall 534.000 mm

0.534 m

j) Effective Depth for Vertical Tensial Reinforcement on Inner Faces of Wall 525.000 mm

0.525 m

k) Effective Depth for Vertical Tensial Reinforcement on Outer Faces of Wall 500.000 mm

0.500 m

l) Strip Width for Horizontal Span (In Vertical Direction), b = 1.000m 1,000.000 mm 1.000 m

m) Strip Width forVertical Span (In Horizontal Direction), b = 1.000m 1,000.000 mm 1.000 m

n) 254.469

o) 254.469

ii) Limits For Maximum Reinforcement, (AASHTO-LRFD-5.7.3.3.1) :.


b) c Variable

c) Variable

Variable

Variable

Variable

Variable mm

Variable mm

Let Provide 14f bars as Vertical Reinforcements on Outer Face of Well. DBar-Ver.-Outer

CCov-Inner

CCov-Outer

de-Inner-H

= tWall - CCov-Inner - DBer-Hor-Inner./2

de-Outer-H

= tWall - CCov-Outer - DBer-Hor-Outer./2

de-Inner-V

= tWall - CCov-Inner - DBer-Hor-Inner - DBer-Ver-Inner./2

de-Outer-V

= tWall - CCov-Outer - DBer-Hor-Outer - DBer-Ver-Outer./2

bH

bV

X-Area of 18f Hori-Bar = p*f18.2/4 Af18, mm2

X-Area of 14f Hori-Bar = p*f14.2/4 Af14. mm2

c/de-Max.











Centroid in mm.







- at Extreme Fiber where Tensile Stress is caused by Externally Applied Forces

- N-mm

Variable

60000000.000

0.06000

60.000/10^3

2.887

c) 173221361.27 N-mm

173.221 kN-m

d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)

Nature of Unfactored Value of Actuat Acceptable M 1.33 Times Maximum

Moment Moment Cracking Factored of M, Allowable Flexural



thus de = ds .



The Cracking Moment of a Section Mcr = Sc(fr + fcpe) - Mdnc(Sc/Snc-1) £ Scfr Mcr








iv) Snc=Section Modulus of Extreme Fiber of the Monolithic or Noncomposite Snc mm3


For the Rectangular RCC Well Section value of Snc = (bHtWall3/12)/(tWall/2) m3








1.2 Times Mr

Mcr-1 Mcr of Mcr


And for the As per Moment Cracking Cracking Moment Factored Min. Moment Moment

Span Section Equation Value Moment Moment of Section Moment for RCC

Direction 5.7.3.3.2-1 M (1.33*M)

kN-m/m kN-m/m kN-m/m kN-m/m kN-m/m kN-m/m kN-m/m kN-m/m kN-m/m22.424 173.221 173.221 173.221 207.866 36.313 48.297 207.866 207.866

Moment in

Horizontal30.155 173.221 173.221 173.221 207.866 49.695 66.094 207.866 207.866

Moment in

Horizontal 13.459 173.221 173.221 173.221 207.866 23.679 31.494 207.866 207.866

Moment in

Vertical 28.580 173.221 173.221 173.221 207.866 41.464 55.147 207.866 207.866

Moment in

Vertical

iv)

a)

0.022

b) 0.016

22 Flexural Design of Horizontal Reinforcement on Inner Faces of Well Wall having Strips Width Directionin Vertical :

i)

a) The Calculated (+) ve Moment occurs at Middle of Well Horizontal 36.313 kN-m/m

36.313*10^6 N-mm/m

value, the Main Reinforcement will be on Inner Face of Wall. 207.866 kN-m/m 207.866*10^6 N-mm/m

b) 207.866 kN-m/m

207.866*10^6 N-mm/m

c) Effective Depth for Horizontal Tensial Reinforcement on Inner Face of Wall 559.000 mm

0.559 m

d) 21.235 mm

e) 1,027.242

f) 247.721 mm,C/C

Mu

Sncfr (Mcr-1£Sncfr) (1.2*Mcr) 1.2Mcr (M ³ Mr)

(+) ve.

(-) ve.

(+) ve.

(-) ve.

Calculations for Balanced Steel Ratio - pb & Max. Steel Ratio - pmax ( AASHTO- 8.16.3) :

Balanced Steel Ratio for Grider Section according to AASHTO-1996-8.16.2.2

pb =0.85*0.85*((f/c/fy)*(599.843/(599.843 + fy))), pb


Provision Horizontal Reinforcements on Inner Face of Well Wall against ( + ) ve. Moment :

(+) MTotal-H-Rock-USD

Span having (+) MTotal-H-Rock-USD <Mr, the Allowable Minimum .For (+) ve

Mr

Since (+) MTotal-H-Rock-USD < Mr, the Minimum Flexural Strength Moment.Thus MU

Mr is the Ultimate DesignMoment MU.

de-Inner-H

= tWall - CCov-Inner - DBer-Hor-Inner./2

With Design Moment MU the value of a = dpro*(1-((1-2MU/)/(b1f/cbdpro

2))1/2) areq.

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-H-req-Inn mm2/m

Spacing of Reinforcement = Af18,*b/As-H-req-Inn sreq


g) 150.000 mm,C/C = 150mm,C/C

h) 1,696.460

i) 0.003

j) 38.966 mm

k) 375.260 kN-m/m

l) Mpro>Mu OK

m) ppro<pmax OK

n) Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, thus the Flexural Design


a) 0.450

b) c 33.121 mm

c) 0.85

d) 0.059


iii) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2.1:

a) 337.734 kN-mwhere; 337.734*10^6 N-mm

b) 375.260 kN-m 375.260*10^6 N-mm

c) f 0.90

d)

Let the Spacing of Reinforcement with 18f bars for the Section spro.

The provided Steel Area with 18f bars having Spacing 150mm,C/C As-H-pro-Inn mm2/m

= Af18,*b/spro

Steel Ratio with provided Steel Area, As-H-pro-Inn/bdpro ppro

With provided Steel Area the value of 'a' = As-H-pro-Inn*fy/(0.b1*f/c*b) apro

Resisting Moment with provided Steel Area = As-pro*fy(de - apro/2)/10^6 Mpro



for Reinforcement on Inner Face of Well Wall in Horizontal Direction is OK.












y(d/s-a/2) + 0.85f/



e)


g) 36.313 kN-m 36.313*10^6 N-mm

h) Mr>MTotal-H-Rock-USD Satisfied

iv) Checking Against Max. Shear Force on Well Wall in Horizontal Span at Bedrock Level.

a) The Maximum Shear Force occurs at Face of Well Wall for Horizontal Span 122.896 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 122.896*10^3 N/m

b)

b-i) 1,000.000 mm


503.100 mm 432.000 mm

b-iii) f 0.90

b-iv) - N

c) 2,641.275 kN/m 2641.275*10^3 N/m

3,827.120 kN/m 3827.120*10^3 N/m

2,641.275 kN/m


Section the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2). In Horizontal Rectangular Section

Width b = bH & ds = de, thus the Nominal Resistance, Mn = Asfy(ds-a/2)

For Fixed Supported & Single Reinforced Rectangular Section with Provided (+)Mn-Mid


Calculated Factored (+) ve. Moment M at Mid Span (+)MTotal-H-Rock-USD


Actual Factored Moment M at Mid Span ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

VU.

Force, VMax = VH-Wall-Rock-USD .= VU


bv is Minimum Width of the Section, here bv = bH, the Design Strip Width. bv.




h = tWell-Wall Thickness of Well Wall.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h



(Sinec the Well Wall is a RCC Structure, thus Vp = 0.

The Nominal Shear Resitance Vn for the Section is the Lesser value of Vn-Well-Wall any of Equations as mentioned in Aritical 5.8.3.3 :




2641.275*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)thus on Innerside of Well Well does not require any Shear Reinforcement.

f)Section does not Require any Shear Reinforcement, thus the Flexural Design of Reinforcement on Inner


a)

Where;

b) 25.259

23.954 kN-m 23.954*10^6 N-mm

1,696.460

559.000 mm the Horizontal Tensile Reinforcement on Inner Face of Well Wall.

c) 167.561











Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Well Wall

Surface of Well Wall in Horizontal Direction is OK.





i) M is Calculated Moment for the Section under Service Limit State (+)MTotal-H-Rock-WSD









17,000.000 N/mm


246.000

d)

e) 2,562.694 N/mm




i)

j)

23 Flexural Design of Horizontal Reinforcement on Outer Faces of Well Wall having Strips Width Direction



of Rectangular Section, dc = DBar/2 + CCov-Inner Since Clear Cover at Inner Face of

Well Wall, CCov-Inner = 50mm & Bar Dia, DBar = 18f ; thus dc = (18/2 + 50)mm


Cover = 50mm.In Well Wall the Tension Bars in One Layer & as per Condition








RCC Well Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.






the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Horizontal for Well Wall on Outer Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.



in Vertical :

i)

a) The Calculated (-) ve Moment occurs at Middle of Well Horizontal 32.705 kN-m/m

32.705*10^6 N-mm/m

value, the Main Reinforcement will be on Outer Face of Wall. 207.866 kN-m/m 207.866*10^6 N-mm/m

b) 207.866 kN-m/m

207.866*10^6 N-mm/m

c) Effective Depth for Horizontal Tensial Reinforcement on Outer Face of Wall 534.000 mm

d) 22.272 mm

e) 1,077.377

f) 236.193 mm,C/C

g) 100.000 mm,C/C = 100mm,C/C

h) 2,544.690

i) 0.005

j) 58.449 mm

k) 526.644 kN-m/m

l) Mpro>Mu OK

m) ppro<pmax OK



a) 0.450

b) c 49.682 mm

Provision Horizontal Reinforcements on Outer Face of Well Wall against ( - ) ve. Moment :

(-) MTotal-H-Rock-USD

Span having (-) MTotal-H-Rock-USD > Mr the Allowable Minimum .For (-) ve

Mr

Since (-) MTotal-H-Rock-USD> Mr, the Minimum Flexural Strength Moment.Thus MU

(-) MH-Well-Bot-USD is the Ultimate DesignMoment MU.

de-Outer-H

= tWall - CCov-Outer - DBer-Hor-Outer./2


2))1/2) areq.

Steel Area required for the Section, As-req. = MU/[ffy(dpro - a/2)] As-H-req-Out. mm2/m

Spacing of Reinforcement = Af18,*b/As-req-Out. sreq


The provided Steel Area with 18f bars having Spacing 100mm,C/C As-H-pro-Out mm2/m

= Af18,*b/spro

Steel Ratio with provided Steel Area, As-H-pro-Out/bdpro ppro

With provided Steel Area the value of 'a' = As-H-pro-Out*fy/(0.b1*f/c*b) apro

Resisting Moment with provided Steel Area = As-pro*fy(d - apro/2)/10^6 Mpro



for Reinforcement on Outer Face Wall in Horizontal Direction is OK.




c) 0.850

d) 0.093



a) 473.979 kN-mwhere; 473.979*10^6 N-mm

b) 526.644 kN-m 526.644*10^6 N-mm

c) f 0.90

d) The Nominal Resistance in a Rectangular Section with One Axis Stress having both Prestressing & Nonprestressing

e)

f) 526.644 kN-mSteel Area against Factored (-) ve.Max. Moments at Partion Faces will have 526.644*10^6 N-mm

g) 32.705 kN-m 32.705*10^6 N-mm

h) Mr>MTotal-H-Rock-USD

Satisfied


a) The Maximum Shear Force occurs at Face of Well Wall for Horizontal Span 122.896 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 122.896*10^3 N/m

b)









y(d/s-a/2) + 0.85f/



Section the Nominal Resistance, Mn = Asfy(ds-a/2) + 0.85f/c(b-bw)b1hf(a/2-hf/2). In a Rectangular Section Structure


For Fixed Supported & Single Reinforced Rectangular Section with Provided Mn-Face

value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored (-) ve. Moment M at Mid Span (-) MTotal-H-Rock-USD


Factored Moment M at face of Span ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

VU.

Force, VMax = VH-Wall-Rock-USD= VU



b-i) 1,000.000 mm


480.600 mm 432.000 mm

b-iii) f 0.90

b-iv) - N

c) 2,523.150 kN/m 2523.150*10^3 N/m

3,655.961 kN/m 3655.961*10^3 N/m

2,523.150 kN/m 2523.150*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied


f)Section does not Require any Shear Reinforcement, thus the Flexural Design of Reinforcement on Outer

bv is Minimum Width of the Section, here bv = bH, the Design Strip Width. bv.






















Surface of Well Wall in Horizontal Direction is OK.



a)

Where;

b) 24.068

32.705 kN-m 32.705*10^6 N-mm

2,544.690

534.000 mm the Horizontal Tensile Reinforcement on Outer Face of Well Wall.

c) 214.187

50.000 mmTension Bar. The Depth is Summation the Clear Cover & Radius of the



17,000.000 N/mm


246.000





i) M is Calculated Moment for the Section under Service Limit State (-)MTotal-H-Rock-WSD







of Rectangular Section, dc = DBar/2 + CCov-Outer.Since Clear Cover at Inner Face

of Well Wall, CCov-Outer = 75mm & Bar Dia, DBar = 18f ; thus dc = (18/2 + 50)mm










d)

e) 1,910.259 N/mm




i)

j)

24 Flexural Design of Vertical Reinforcement on Inner Faces of Well Wall having Strips Width Directionin Horizontal :

i)

a) The Calculated (+) ve Moment at Bedrock Level occurs at Middle in 23.679 kN-m/m

23.679*10^6 N-mm/m

.For (+) ve value, the Main Reinforcement will be on Inner Face of Wall. 207.866 kN-m/m 207.866*10^6 N-mm/m

b) 207.866 kN-m/m

207.866*10^6 N-mm/m

c) Effective Depth for Vertical Tensile Reinforcement on Inner Face of Wall 525.000 mm

0.525 m

d) 22.671 mm

e) 1,096.67

f) 232.038 mm,C/C

g) 150.000 mm,C/C = 150mm,C/C

h) 1,696.460


Well Wall Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.






the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Horizontal for Well Wall on Outer Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.


Provision Horizontal Reinforcements on Inner Face of Well Wall against (+) ve. Moment at Bedrock Level:


between Pertition Walls having (+) MTotal-V-Rock-USD <Mr, the Allowable Minimum

Mr

Since (+) MTotal-V-Rock-USD < Mr, the Minimum Flexural Strength Moment.Thus MU

Mr is the Ultimate DesignMoment MU.

de-Inner-V

= tWall - CCov-Inner - DBer-Hor-Inner - DBer-Ver-Inner./2


2))1/2) areq.

Steel Area required for the Section, As-req. = MU/(ffy(dpro - a/2)) As-V-req-Inn mm2/m

Spacing of Reinforcement = Af14,*b/As-V-req-Inn sreq


The provided Steel Area with 14f bars having Spacing 150mm,C/C As-V-pro-Inn mm2/m


i) 0.001

j) 38.966 mm

k) 351.612 kN-m/m

l) Mpro>Mu OK

m) ppro<pmax OK



a) 0.450

b) c 33.121 mm

c) 0.85

d) 0.063



a) 316.450 kN-mwhere; 316.450*10^6 N-mm

b) 351.612 kN-m 351.612*10^6 N-mm

c) f 0.90

d) The Nominal Resistance for a Flanged Section with One Axis Stress having both Prestressing & Nonprestressing

e)

= Af14,*b/spro

Steel Ratio with provided Steel Area, As-V-pro-Inn/bdpro ppro

With provided Steel Area the value of 'a' = As-V-pro-Inn*fy/(0.b1*f/c*b) apro

Resisting Moment with provided Steel Area = As-pro*fy(de - apro/2)/10^6 Mpro



for Reinforcement on Inner Face Wall in Vertical Direction is OK.











y(d/s-a/2) + 0.85f/




Width b = bV & ds = de, thus the Nominal Resistance, Mn = Asfy(ds-a/2)



g) 23.679 kN-m 23.679*10^6 N-mm

h) Mr>MTotal-H-Rock-USD Satisfied

iv) Checking Against Max. Shear Force on Well Cap in X-X Direction.

a) The Maximum Shear Force occurs at Bedrock of Well Wall in Vertical Span, 171.876 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 171.876*10^3 N/m

b)

b-i) 1,000.000 mm


472.500 mm 432.000 mm

b-iii) f 0.90

b-iv) - N

c) 2,480.625 kN/m 2480.625*10^3 N/m

3,594.343 kN/m 3594.343*10^3 N/m

2,480.625 kN/m 2480.625*10^3 N/m


For Fixed Supported & Single Reinforced Rectangular Section with Provided (+)Mn-Mid


Calculated Factored (+) ve. Moment M at Mid Span (+)MTotal-V-Rock-USD


Actual Factored Moment M at Mid Span ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

VU.

Force, VMax = VV-Wall-Rock-USD .= VU


bv is Minimum Width of the Section, here bv = bV, the Design Strip Width. bv.













c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied

e)thus on Innerside of Well Well does not require any Shear Reinforcement.

f)Section does not Require any Shear Reinforcement, thus the Flexural Design of Reinforcement on Inner


a)

Where;

b) 18.471

16.451 kN-m 16.451*10^6 N-mm

1,696.460

525.000 mm the Horizontal Tensile Reinforcement on Inner Face of Well Wall.

c) 142.791

75.000 mmTension Bar. The Depth is Summation of Clear Cover & Radius of the











Surface of Well Wall in Vertical Direction is OK.





i) M is Calculated Moment for the Section under Service Limit State (+)MTotal-V-Rock-WSD







of Rectangular Section, dc = DBar-H + DBar-V/2 + CCov-Inner Since Clear Cover at




17,000.000 N/mm


246.000

d)

e) 2,199.092 N/mm




i)

j)

25 Flexural Design of Vertical Reinforcement on Outer Faces of Well Wall having Strips Width Directionin Horizontal :

i)

Inner Face of Well Wall, CCov-Inner = 50mm & Bar Dia, DBar-H = 18f and

DBar-V = 14f thus dc = (18 +14/2 + 50)mm
















the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Vertical for Well Wall on Inner Surface in respect ofControl of Cracking & Distribution of Reinforcement are OK.


Provision Vertical Reinforcements on Outer Face of Well Wall against ( - ) ve. Moment :


a) The Calculated (-) ve Moment occurs at Middle of Well Horizontal 41.464 kN-m/m

41.464*10^6 N-mm/m

value, the Main Reinforcement will be on Outer Face of Wall. 207.866 kN-m/m 207.866*10^6 N-mm/m

b) 207.866 kN-m/m

207.866*10^6 N-mm/m

c) Effective Depth for Vertical Tensile Reinforcement on Inner Face of Wall 500.000 mm

0.500 m

d) 23.860 mm

e) 1,154.181

f) 220.476 mm,C/C

g) 100.000 mm,C/C = 100mm,C/C

h) 2,544.690

i) 0.005

j) 58.449 mm

k) 491.171 kN-m/m

l) Mpro>Mu OK

m) ppro<pmax OK



a) 0.450

b) c 49.682 mm

c) 0.85

(-) MTotal-V-Rock-USD

Span having (-) MTotal-V-Rock-USD > Mr the Allowable Minimum .For (-) ve

Mr

Since (-) MTotal-V-Rock-USD> Mr, the Minimum Flexural Strength Moment.Thus MU

(-) MTotal-V-Rock-USD is the Ultimate DesignMoment MU.

de-Outer-V

= tWall - CCov-Outer - DBer-Hor-Outer - DBer-V-Outer./2


2))1/2) areq.

Steel Area required for the Section, As-req. = MU/[ffy(dpro - a/2)] As-V-req-Out. mm2/m

Spacing of Reinforcement = Af14,*b/As-V-req-Out. sreq


The provided Steel Area with 14f bars having Spacing 100mm,C/C As-V-pro-Out mm2/m

= Af14,*b/spro

Steel Ratio with provided Steel Area, As-V-pro-Out/bdpro ppro

With provided Steel Area the value of 'a' = As-V-pro-Out*fy/(0.b1*f/c*b) apro

Resisting Moment with provided Steel Area = As-pro*fy(d - apro/2)/10^6 Mpro



for Reinforcement on Outer Face Wall in Vertical Direction is OK.






d) 0.099



a) 442.054 kN-mwhere; 473.979*10^6 N-mm

b) 491.171 kN-m 491.171*10^6 N-mm

c) f 0.90

d) The Nominal Resistance in a Rectangular Section with One Axis Stress having both Prestressing & Nonprestressing

e)

f) 491.171 kN-mSteel Area against Factored (-) ve.Max. Moments at Bedrock Faces will have 491.171*10^6 N-mm

g) 41.464 kN-m 41.464*10^6 N-mm

h) Mr>MTotal-V-Rock-USD

Satisfied


a) The Maximum Shear Force occurs at Bedrock of Well Wall for Vertical Span 171.876 kN/mwhich is also Ultimate Shearing Force for the Section. Thus Maximum Shear 171.876*10^3 N/m

b)

b-i) 1,000.000 mm








y(d/s-a/2) + 0.85f/





For Fixed Supported & Single Reinforced Rectangular Section with Provided Mn-Face

value of Nominal Resistance, Mn = Asfy(ds-a/2)

Calculated Factored (-) ve. Moment M at Bedrock Face of Span (-) MTotal-V-Rock-USD


Factored Moment M at face of Span ( Which one is Greater, if Mr ³ M the Flexural Design for the Section has Satisfied otherwise Not Satisfied)

VU.

Force, VMax = VV-Wall-Rock-USD= VU


bv is Minimum Width of the Section, here bv = bV, the Design Strip Width. bv.



450.000 mm 432.000 mm

b-iii) f 0.90

b-iv) - N

c) 15,125.597 kN/m 15125.597*10^3 N/m

15,125.597 kN/m 15125.597*10^3 N/m

46,125.000 kN/m 46125.000*10^3 N/m


c-ii) 0.000 N/m

c-iii) b 2.000

c-iv) 0.000 N

d) Vn>Vu Satisfied


f)Section does not Require any Shear Reinforcement, thus the Flexural Design of Reinforcement on Outer


a)





















Surface of Well Wall in Vertical Direction is OK.




Where;

b) 26.382

33.567 kN-m 33.567*10^6 N-mm

2,544.690

500.000 mm the Horizontal Tensile Reinforcement on Outer Face of Well Wall.

c) 163.455

75.000 mmTension Bar. The Depth is Summation of Clear Cover & Radius of the



17,000.000 N/mm


246.000

d)



i) M is Calculated Moment for the Section under Service Limit State (-)MTotal-V-Rock-WSD







of Rectangular Section, dc = DBar-H + DBar-V/2 + CCov-Outer Since Clear Cover at

Outer Face of Well Wall, CCov-Outer = 75mm & Bar Dia, DBar-H = 18f and

DBar-V = 14f thus dc = (18 +14/2 + 50)mm










Well Wall Structure, thus value of the Crack Width Parameter Z should calculate based the value of fs-Dve.


e) 2,743.875 N/mm




i)

j)

26 Design of Well Wall under Provisions of Compression Member :

i) Dimensions of Assumed Rectangular Column Strip of Well Wall & Related Features:

a) Width of Assumed Column Strip = 1.000m b 1.000 m

b) h 0.600 m

c) I 18000000000.0

0.018

d) 4.825 m

e) N 324.433 kN/m

f) Factored Horizontal Load At Top of Well Due to Soil & Surcharge 42.531 kN/m

g) 1,500.000

h) T 3.100

i) 12.400 m

ii) Calculation of Moment due to Applied Horizontal Forces :

a) H 42.531 kN/m

b) Moment due to Max. Horizontal Pressure with Scour Depth M 221.266 kN-m/m






the Computed Tensile Stress fsa < 0.6fy ;the Developed Crack Width Parameter ZDev. < ZMax. Allocable Max. Crack Width Parameter, thus Provisions of Tensile Reinforcement in Vertical for Well Wall on Outer Surface in respect of Control of Cracking & Distribution of Reinforcement are OK.


Depth of Assumed Column Strip = tWell-Wall = 0.600m

Moment of Inertia of Assumed Column Strip = bh3/12 mm4

m4

Assumed Depth of Scour = HWell-pro. - HB-Rock dScour

Factored Vertical Load upon Assumed Column Strip = pStaining-USD

PH-Well-Top.-USD

Equally Carried by Wells Rear & Front Walls = åFPH-Well-Top.-USD/2

Coefficient of Horizontal Subgreade Modulus of Soil = ήh ήh kN/m3

Stiffness Factor T = (EcI/ήh)0.2

Minimum Depth required Well Wall = 4* T Ld-req

Maximum Horizontal Force on Assumed Column Strip = pH-Y-Y-USD


iii) Events Related to Calculation of Core Width :

a) Provided Diameter of Horizontal Reinforcements on Inner Face of Well. 18 mm

b) Provided Diameter of Horizontal Reinforcements on Outer Face of Well. 18 mm

c) Provided Diameter of Vertical Reinforcements on Inner Face of Well. 14 mm

d) Provided Diameter of Vertical Reinforcements on Outer Face of Well. 14 mm

e) Provided Clear Cover of Well Walls on Inner Face, 50 mm

f) Provided Clear Cover of Well Walls on Outer Face, 75 mm

g) d 425.000 mm 0.425 m

h) d/h 0.71

i) 153.938

iii) Events Related to Provision of Compression Reinforcements Using UK Code-BS8110:

a) 0.03

b) 0.03

iv ) Calculations for Reinforcement by Using the BS8110 British Code of Practic Data Sheet 53 for d/h = 80 &

(Example of the Design of Reinforced Concrete Buildings).

iv-i)

a) 0.0001

b) 0.0002

c) 0.0002

= 0.0001 + (410-250)*(0.0002-0.0001)/(460-250)

= 0.5*(1.8*T +dScour) + H

DBar-H-Inner

DBar-H-Outer

DBar-V-Inner

DBar-V-Outer

CCov-Inner

CCov-Outer

Core Width of Well Wall = tWall - CCov-Inner - CCov-Outer -2*DBer-H.+ DBer-V.

Ratio between Core Width of Wall & Total Thickness of Wall = d/h

X-Sectional Area of 14f Bar = p*D-f-142/4 Af-14 mm2

For Assumed Rectangular Column Section tne Value of N/bhSf/c N/bhf/

c

For Assumed Rectangular Column Section tne Value of N/bhSf/c M/bh2f/

c

Sheet 54 for d/h = 75 for Calculation of value of p1/f/c & computing same for d/h = 0.71 :

Compution of Data Sheet 53 for Value of fy = 410 kN/mm2

With values fy = 250N/mm2, d/h = 0.80, M/bh2f/c = 0.03,N/bhf/

c = 0.03 p1/f/c

the value of p1/f/c = 0.0000+(0.0005-0.0000)*5/17

With values fy = 460N/mm2, d/h = 0.80, M/bh2f/c = 0.03, N/bhf/

c = 0.03 p1/f/c

the value of p1/f/c = 0.0000+(0.0005-0.0000)*4/13

Thus form values a & b for fy =410N/mm2,with d/h =0.80, M/bh2f/c =0.03 & p1/f/

c

N/bhf/c = 0.03 the Commutated value of p1/f/

c


iv-ii)

a) 0.0001

b) 0.0000

c) 0.0001

= 0.0000+ (410-250)*(0.0001-0.0000)/(460-250)

iv-iii)

a)

b)

c) 0.0001

iv-iv) Calculations for Reinforcements Considering Well Wall as Compression Component :

a) 0.002

b) 1,260.000

c) Steel Area Provided for the Rectangular Section as Vertical Reinforcements 4,241.150

d) Relation between Required Compression Component Steel Area & Provided As-Comp.<As-Total-Flx. Satisfied.Flexural Component Steel Area (Whether Requirement of Compression Component is being Satisfied).

e) Since the Provision of Vertical Reinforcements of Well Wall as Flexural Component is Higher than that Required as Compression Component, thus the Structue is Safe as a Compression Component also.

27 Checking for Provisionsof Vertical Reinforcements of RCC Well Wall as Compression Component :

i) Philosophy in Designing Vertical Reinforcements for RCC Well as Compression Member & Checking:

a)It will also have to Receive Horizontal Loads those will Imposed upon Well's Outer Surfaces due to Lateral Soil & Surcharge Pressure in between Well Cap Bottom & Bed-rock Top during Construction. After Construction Inner

Compution of Data Sheet 54 for Value of fy = 410 kN/mm2

With values fy = 250N/mm2, d/h = 0.75 M/bh2f/c = 0.03,N/bhf/

c = 0.03 p1/f/c

the value of p1/f/c = 0.0000+(0.0005-0.0000)*4/18

With values fy = 460N/mm2, d/h = 0.75, M/bh2f/c = 0.03r, N/bhf/

c = 0.03 p1/f/c

the value of p1/f/c = 0

Thus form values a & b for fy =410N/mm2,with hs/h =0.75, M/h3f/c =0.09 & p1/f/

c

Nmin/h2f/c = 0.09 the Commutated value of p1/f/

c

Compution of Value of p1/f/c againstd/h = 0.71 for fy = 410 N/mm2 using the Computed value of p1/f/

c

Computed Values of d/h = 0.80 & 0.75 for fy = 410N/mm2:

Against d/h = 0.80 & fy = 410N/mm2 p1/f/c = 0.0002

Against d/h = 0.75 & fy = 410N/mm2 p1/f/c = 0.0001

Thus for d/h =0.71 & fy =410N/mm2 p1/f/c = 0.0001+(0.0002-0.0001)*9/5 p1/f/

c

With Commutated value of p1/f/c = 0.0001, p1 =0.0001*f/

c p1

For fy = 410N/mm2, f/c = 21N/mm2, d/h = 0.71, & p1 = 0.002 for Rectangular As-Comp. mm2

Section the Required Total Steel Area for both Faces Wall, As = p1*b*h

As-Total-Flx. mm2

under Flexural Arrangement for both Faces = As-V-pro-Inn + As-V-pro-Out

The RCC Well will have to Receive all Vertical Load Reactions (DL & LL) from Superstructure & Substructures.


Pockets of Well will be filled up by Bottom Plugging, Filling Material & Finally the Top Plugging. After completionthe Outer Horizontal Pressures will be Balanced by the Lateral Pressures of filled Inner Materials.

ii) Checking of Vertical Reinforcements for RCC Well inrespect of Compression Member :

a) The Provided Vertical Reinforcement for RCC Wells should not be less than 0.20% of the Actual Grosse Cross-Sectional Area of Staining & at least of which One-third (33%) on Inner Face.(Concrete Bridge Practice : Analysis, Design and Economice- RAINA; The Substructure Pager-79).

b) Min of Steel Area Required as Vertical Reinforcement on the Staining Wall 0.200 %as % of Grosse Cross-Sectional Area.

c) Min of Steel Area Required on Inner Face of the Staining Wall in respect Total 33.000 %Total Steel Area as Vertical Reinforcement for of Staining.

d) Provided Steel Area as Vertical Reinforcement for per meter Length of the 1,696.460

e) Provided Steel Area as Vertical Reinforcement for per meter Length of the 2,544.690

d) Provided Total Steel Area as Vertical Reinforcement for per meter Length 4,241.150

e) Percentage of Steel Area Provided as Vertical Reinforcement against Grosse 0.707 %Crosse-sectional Area for 1 (One) meter Length of of Staining of Well.

f) Whether the Provided Total Steel Area against Vertical Reinforcement have Satisfied Provision Satisfied the Minimum Reqirment or Not.

g) Percentage of Steel Area on Inner Face against Total Provided Steel Area as 40.000 %

h) Whether the Provided Steel Area against Vertical Reinforcement on Inner Face Satisfied Provision Satisfied the Minimum Reqirment or Not.

g) Since the Percentage of Steel Area Provided as Vertical Reinforcement against Grosse Crosse-sectional Area of Staining of Well have Satisfied the Minimum Requirment & also the Steel Area on Inner Face of the Staining haveSatisfied the Minimum Requirement against Total Provided Steel Area as Vertical Reinforcement, thus Arranges for

28 Checking of Provided Horizontal Reinforcements for RCC Well Wal in Respect of Hoop Tension :

i) Provisions of Hoop Tension for Horizontal Reinforcements in Staining of Well :

%Min.-Gross

%Min.Inner

As-V-pro-Inn. mm2/m

Staining of Well on Inner Face = As-V-pro-Inn.mm2

As-V-pro-Out. mm2/m

Staining of Well on Outer Face = As-V-pro-Out.mm2

As-V-pro-Total. mm2/m

of Staining of Well (Both Faces) = As-V-pro-Inn.+ As-V-pro-Out.

%As-V

= 100*As-V-pro-Total/b*tWall

%As-V-Inn.

Vertical Reinforcement on Staining of Well = 100*As-V-pro-Inn./As-V-pro-Total.

Vertical Reinforcement is OK.


a) The Staining of RCC Well should Provide Hoop Reinforcement at lest 0.04% by Volume against Unit Length of Staining. (Concrete Bridge Practice : Analysis, Design and Economice- RAINA; The Substructure Pager-79).

b) Minimum Requirment as Hoop Reinforcement by Volume for Unit Horizontal 0.040 %Length of Staining of Well.

c) Provided Steel Area as Horizontal Reinforcement for per meter Height 1,696.460

d) Provided Steel Area as Horizontal Reinforcement for per meter Height 2,544.690

e) Provided Total Steel Area as Horizontal Reinforcement for per meter 4,241.150

f) Required Steel Area as Horizontal Reinforcement as per Calculation for per 1,027.242

g) Required Steel Area as Horizontal Reinforcement as per Calculation for per 1,077.377

h) Required Total Steel Area as per Calculation as Horizontal Reinforcement 2,104.619

i) Additional Horizontal Steel Provided against Calculated Steel for per meter 2,136.531

j) Grosse Volume of Concreta for 1 (One) meter Length & 1 (One) Height of 600000000.000

k) Provided Addl. Steel Volume in 1 (One) meter Length & 1 (One) Height of 2136531.216

l) Percentage of Addl. Steel Volume Provided as Horizontal Reinforcement 0.356 %

m) Whether the Provided Addl. Steel Volume as Horizontal Reinforcement for Unit Length Provision Satisfied of Staining have Satisfied the Minimum Reqirment of Hoop Reinforcement or Not.

n) Since the Addl. Steel Volume Provided as Horizontal Reinforcement of Staining of Well have Satisfied the MinimumRequired Volume of Steel as Hoop Reinforcement, thus Design of Staining of Well for Reinforcement in Respect of

Vs-Hoop.

As-H-pro-Inn. mm2/m

of Staining of Well on Inner Face = As-H-pro-Inn.mm2

As-H-pro-Out. mm2/m

of Staining of Well on Outer Face = As-H-pro-Out.mm2

As-H-pro-Total. mm2/m

Height of Staining of Well (Both Faces) = As-H-pro-Inn.+ As-H-pro-Out.

As-H-req-Inn. mm2/m

meter Height of Staining of Well on Inner Face = As-H-req-Inn.

As-H-req-Out. mm2/m

meter Height of Staining of Well on Outer Face = As-H-req-Out.

As-H-req-Total. mm2/m

for per meter Height of Staining of Well (Both Faces) = As-H-req-Inn.+ As-H-req-Out.

As-H-pro-Addl. mm2/m

Length of Staining of Rcc Well (Both Faces) = As-H-pro-Total.- As-H-req-Total.

VStain. mm3/m

Staining of Well, VStain. = 1000*1000*tWell.

VH-Steel-Addl. mm3/m

Staining of Well, VH-Steel-Addl. = 1000*As-H-pro-Addl. mm3

%H-Steel-Addl.

= 100*VH-Steel-Addl./VStain.

Hoop Tension is OK.


Mr)


Satisfied


Satisfied.

Q. Design of Elastomeric Bearings for Abutment :

1 Phenomenon in Designing of Elastomeric Bearing for Simple Supported Bridge Structure :

i) Provisions of AASHTO-LRFD- Section-14(SI) : Joints And Bearings in Respect of Design of Steel ReinforcedElastomeric Bearings for Bridge Structures (Claues 14.4.1;:

a) The Bearings should be Designed to allow the Deformations due to Temperature, Creep, Shrinkage and other TimeBased causes. Those should be consistence with proper Functioning of the Bridge Structure

b) The Bearings should be Designed to resist Applied Loads & Accommodate Movements under Service Limit State &Strength Limit State. Those Should also Satisfy thru requirements of Fatigue & Fracture Limit States. No Damagedue Movement of Bearing Under Service Limit State can occur. No Irreparable Damage Under Strength Limit State & Extreme Limit State is Allowable.

c) All Critical Load Combinations should considered for Designing of Bearings in Respect of Translational & Rotational . Movement of Bridge Structure. In these Cases Three-Dimensional Effect should be Considered. In these Respect for both Translational & Rotation should be Considered about Two Horizontal Axis and also about the Vertical Axis.

d) Both Instantaneous & Long-term Effect should be Considered in Design of Bearings. Influence of Impact or Dynamic

e)

f) Calculated Maximum Rotation for Bearing shall be the Total Rotation due to Unfactored Service Loads. This will have an Allowance of 0.005 Red.

g)should Followed in Respect of Thermal Movement with Uniform Temperature Change.

2 Sketch Diagram of Elastomeric Bearing, Dimensions, Applied Load & Related Data :

i) Sketch Diagram of Proposed Elastomeric Bearing Sections :

a) Longitudinal Section in Elevation Showing 50% Length :

h

Load Allowance (IM) should not be included in Design of Bearing.

The Minimum Thermal Movement shall be Computed from the Extreme Temperature According to Article 3.12.2 & the Design Loads shall be based upon Load Combination & Load Factotrs as Mentioned in Section-3.

In Design of Bearings Either of the Procedure/Method-A or Procedure/Method-B as Mentioned in Article-3.12.2

tEla-Out

ts-Out

tEla-Inn

ts-Inn

tEla-Inn

ts-Inn

tEla-Inn

ts-Inn

tEla-Inn

L/2

b) Longitudinal Plan Showing 100% Length & Width:

Traffic

L = Length

ii) Superstructure Related Data :

a) Span Length of Girgers (Distance between Bearing C/C) 24.400 m

b) Number of Main Girders of Bridge Structure 5.000 nos.

c) Factored Dead Load Reaction from Superstructure for One No. Interior Girder 512.665 kN

d) Factored Live Load Reaction from Superstructure for One No. Interior Girder 636.813 kN

e) Factored Dead Load Reaction from Superstructure for One No. Interior Girder 436.375 kN

f) Factored Live Load Reaction from Superstructure for One No. Interior Girder 257.965 kN

Diflection of Bridge Girder at Mid Spam

g) 40.881 mm

39.823 mm

h) Total Rotation at Support for Bridge Girder

tS-Out

tEla-Out

CCov.

CCov.

SL-C/C

NGir.

RDL-USD

under Strength Limit State (USD) Provisions.

RLL-USD

under Strength Limit State (USD) Provisions.

RDL-WSD

under Service Limit State (WSD) Provisions.

RLL-WSD

under Service Limit State (WSD) Provisions.

i) Under Strength Limit State (USD) DUSD

ii) Under Service Limit State (WSD) DWSD

W =

Wid

th

2.449E-15 Rad.

1.588E-15 Rad.

1.588E-15 Red.

iii) Dimensions of Proposed Elastomeric Bearing :

a)Thickness of Top & Bottom Cover Layers should not be Thicker than 70% that of Internal Layers.

b) Width of Bridge Main Girder 350.000 mm

c) Transverse Width of Elastomeric Bearing 300.000 mm

d) Longitudinal Length of Elastomeric Bearing 400.000 mm

e) Thickness of Internal Elastomeric Layers 12.000 mm

f) Thickness of Outer or Cover Elastomeric Layers 4.500 mm

g) Thickness of Outer Face Steel Plate Layers 3.000 mm

h) Thickness of Inner Steel Plate Layers (2 nos. Plate each of 1.500mm thick) 3.000 mm

i) Clear Cover on All Sides of Bearing 3.000 mm

j) Total Number of Internal Elastomeric Layers 4.000 nos.

k) Total Number of Internal Steel Plate Layers 3.000 nos.

l) Total Thickness of Elastomeric in Bearing 57.000 mm

m) Total Thickness of Steel Plats in Bearing 15.000 mm

n) 72.000 mm

o) Relation between Cover Elastomeric Layer & Internal Elastomeric Layer in % 37.500 %

p) Max. Allowable % of Cover Elastomeric with that of Tnternal Elastomeric 70.000 %

q) Whether Provisons for Thicknees of Cover Layer & Internal Layer have Satisfied or Not. Satisfied

iv) Loads upon Elastomeric Bearing under Strength Limit State (USD):

i) Under Strength Limit State (USD) qUSD

ii) Under Service Limit State (WSD) qWSD

iii) Under Service Limit State (WSD) due to Live Loads onlu. qUFL

According to AASHTO-LRFD-14.57.5.1. All Internal Layers of Elastomer should be of Same Thickness. Whereas

bGir.

WBear.

LBear.

tEla-Inn

tEla-Out

ts-Out.

ts-Inn.

CCov.

NEla-Inn.

Ns-Inn.

HEla.

= NEla-Inn*tEla-Inn + 2*tEla-Out

HS

= NS-Inn*tS-Inn + 2*tS-Out

Total Thickness of Elastomeric Bearing = HEla + HS HBearing

%tEla-Out

= 100*tEla-Out/tEla-Inn

%tEla-Out-Allow.

a) Elastomeric Bearings have to face Minimum Loading when there will be only 636.813 kN 636.813*10^3 N

b) Elastomeric Bearings have to face Maximum Loading when there will be both 1,149.477 kN 1213.748*10^3 N

v) Loads upon Elastomeric Bearing under Strength Limit State (USD):

a) Elastomeric Bearings have to face Minimum Loading when there will be only 257.965 kN 357.433*10^3 N

b) Elastomeric Bearings have to face Maximum Loading when there will be both 694.340 kN 810.356*10^3 N

vi) Shape Factor for Elastomeric Bearing Layer :

a) 7.143

b) L 400.000 mm

c) W 300.000 mm

d) 12.000 mm

vi) Properties of Elastomeric & Steel :

a) Difference of Atmosphere Temperature of the Locality for Bridge Structure T 35 0

b) G 0.950

c) 290.816 Mpa

e) 410.00 MPa

f) a 0.000011

g) Shrinkage Strain of Concrete 0.0003

PMin.-USD-LL

Dead Loads (DL) of the Bridge Superstructure under Strength Limit State (USD).

PMax.-USD

Dead Loads (DL) of the Bridge Superstructure & Live Loads (LL) upon it under Strength Limit State (USD)

PMin.-WSD-LL

Dead Loads (DL) of the Bridge Superstructure under Service Limit State (WSD).

PMax.-WSD

Dead Loads (DL) of the Bridge Superstructure & Live Loads (LL) upon it under Service Limit State (WSD)

According to Equation -14.7.5.1-1 (AASHTO-LRFD) the Shape Factor for Si.

Layer of Elastomeric Bearing is Si = LW/2hri(L+W); where

L is Length of Rectangular Elastomeric Bearing along Bridge Longitudinal

Axis in mm (Parallel to Traffic) = Blearing.

W is Width of Bearing in Transverse Direction in mm = Wbear

hri is Thickness of Internal Elastomeric Layer in mm = tEla-Inn hri.

OC(AASHTO-LRFD-3.12.2.1.1-Procudere-A ; Table 3.12.2.1.1-1)

According to AASHTO-LRFD-14.7.5.2 the Shear Modulus of Elastomer forHardness Shore-A having value 50 to 60 will be in between 0.60 to 1.30MPa.

Let Consider the Average of .60 & 1.30 MPa at 27OC of Average Temperature. (Table 14.7.5.2-1)

According to AASHTO-LRFD; Equ-C14.6.3.2-1 the Effective Compressive Ec.

Modulus of Elastomeric Bearing Ec = 6*GSi2 in Mpa. Here,


Co-efficient of Thermal Expansion of Normal Density Concrete = 10.8*10-6/OC mm/mm/OC(AASHTO-LRFD-5.4.2.2)

DSH

3 Checking for Compressive Stress on Elastomeric Bearing under Provision of AASHTO-LRFD-14.7.5.3.2 :

i) Allowable Compressive Shear Stress for Elastomeric Bearing under Service Limit State (WSD) :

a) The Allowable Shear Stress for Bearing Subject to Shear Deformation due to 11.264 Mpa

11.000 Mpa

b) The Allowable Shear Stress for Bearing Subject to Shear Deformation due to 4.479 Mpa

c) The Allowable Shear Stress for Bearing Fixed Against Shear Deformation 13.571 Mpa

12.000 Mpa

d) The Allowable Shear Stress for Bearing Fixed Against Shear Deformation 6.786 Mpa

ii) Allowable Plan Area for Elastomeric Bearing under Service Limit State (WSD) :

a) 61,640.825

b) 57,599.811

c) 51,161.885

d) 38,015.875

iii) Checking for Provided & Requuired Plan Area of Elastomeric Bearing for Compressive Shear Stresses :

a) 120000.000

b) Apro-Plan.>AsS-req. OKCompressive Shear Stress under Shear Deformation for Total Load:

c) Apro-Plan.>AsL-req. OKCompressive Shear Stress under Shear Deformation for Live Load only:

c) Apro-Plan.>AsS-Fix-req. OKfor Compressive Shear Stress under Shear Deformation for Total Load:

sS-Cal.

Total Load (DL + LL) is Expressed by sS £ 1.66GS £11.00MPa. sS-Allow. (Equation-14.7.5.3.2-1)

sL-Allow

Live Load (LL) only is Expressed by sL £ 0.66GS MPa. (Equation-14.7.5.3.2-2)

sS-Fix-Cal

due to Total Load (DL + LL) is Expressed by sS £ 2.00GS £12.00MPa. sS-Fix-Allow

(Equation-14.7.5.3.2-3)

sL-Fix-Allow

due to Live Load (LL) Expressed by sL £ 1.00GS MPa. (Equation-14.7.5.3.2-4)

Required Plan Area of Elastomeric Bearing against Equation-14.7.5.3.2-1 AsS-req. mm2

= PMax.-WSD/sS

Required Plan Area of Elastomeric Bearing against Equation-14.7.5.3.2-2 AsL-req. mm2

= PMin.-WSD-LL/sL

Required Plan Area of Elastomeric Bearing against Equation-14.7.5.3.2-3 AsS-Fix-req. mm2

= PMax.-WSD/sS-Fixed

Required Plan Area of Elastomeric Bearing against Equation-14.7.5.3.2-5 AsL-Fix-req. mm2

= PMin.-WSD-LL/sL-Fixed

Provided Plan Area of Elastomeric Bearing = L*W Apro-Plan mm2

Relation between Provided Area & Required Area of Elastomeric Bearing for

Relation between Provided Area & Required Area of Elastomeric Bearing for

Relation between Provided Area & Required Area of Elastomeric Bearing

d) Apro-Plan.>AsL-Fix-req. OKfor Compressive Shear Stress under Shear Deformation for Live Load only:

iv) Developed Compressive Shear Stress for Elastomeric Bearing under Service Limit State (WSD) :

a) Developed Compressive Shear Stress on Elastomeric Bearing due to Applied 5.786 Mpa

b) Developed Compressive Shear Stress on Elastomeric Bearing due to Applied 2.150 Mpa

c) Relation between Developed Compressive Stress on Elastomeric Bearing and sS-Dev.<sS-Allow OKAllowable Compressive Stress for Bearing Subject to Shear Deformation due to

d) Relation between Developed Compressive Stress on Elastomeric Bearing and sL-Dev.<sL-Allow OKAllowable Compressive Stress for Bearing Subject to Shear Deformation due to

e) Relation between Developed Compressive Stress on Elastomeric Bearing and sS-Dev.<sS-Fix-Allow OKAllowable Compressive Stressfor Bearing Fixed Against Shear Deformation due to

f) Relation between Developed Compressive Stress on Elastomeric Bearing and sL-Dev.<sL-Fix-Allow OKAllowable Compressive Stressfor Bearing Fixed Against Shear Deformation due to

n) Since the Developed Compressive Stresses on Elastomeric Bearing are Less than All the Allowable Limits& the Provided Plan Area of Bearing has Satisfied the Minimum Reqired Areas for the Respective Stresses,thus both in respect of Plan Area & Compressive Stress Proposed the Elastomeric Bearing is OK .

4 Checking for Compressive Deflection on Elastomeric Bearing under Provision of AASHTO-LRFD-14.7.5.3.3 :

i) Allowable Compressive Deflection on Elastomeric Bearing under Service Limit State (WSD) :

a)

b)

c)Against Respective Application of Load/Loads.

d) 12.000 mm

e) 3.400 %Graph the % of Strain for Allowable Stress for Bearing due to Developed Max.

Relation between Provided Area & Required Area of Elastomeric Bearing

sS-Dev.

Total Loads (DL + LL) = PMax.-WSD/(L*W)

sL-Dev.

Live Load (LL) only = PMin.-WSD-LL/(L*W)

Applied Total Loads (DL + LL) under Service Limit State (WSD).

Applied Live Loads (LL) only under Service Limit State (WSD).

Applied Total Loads (DL + LL) under Service Limit State (WSD).

Applied Live Loads (LL) only under Service Limit State (WSD).

The Deflection in Elastomeric Bearing should be Calculated both for Total Applied Loads (DL + LL) an also for theApplied Live Load (LL) Separately.

The Instantaneous Deflection in Elastomeric Bearing is Expressed by the Equ. 14.7.5.3.3-1; d = Sεihri ; where,

εi is Instantaneous Compressive Strain in Total Internal Elastomeric Layers, εi

hri is Thickness of Each Internal Elastomeric Layers in mm = tEla-Inn hri.

Using the Figure C-14.7.5.3.3-1 with 50 Durometer Reinforced Bearing %εi

f) Thus the Instantaneous Compressive Strain in Elastomeric Bearing due to 0.034

g) The Allowable Instantaneous Deflection in Elastomeric Bearing for Total

0.408 mm

ii) Developed Compressive Deflection on Elastomeric Bearing under Service Limit State (WSD):

a) Calculated Developed Compressive Deflection of Elastomeric Bearing due to 0.239 mm

b) Calculated Developed Compressive Deflection of Elastomeric Bearing due to 0.089 mm

c) Relation between Developed Compressive Deflection of Elastomeric Bearing and dDev-S.<dAllow OK

d) Relation between Developed Compressive Deflection of Elastomeric Bearing and dDev-L.<dAllow OK

e) Since in both the for Total Loads & Live Load only the Developed Compressive Deflections of Elastomeric Bearing are within Allowable Limit, thus in Respect of Compressive Deflection Proposed Bearing is OK.

5 Checking for Shear Deformation of Elastomeric Bearing under Provision of AASHTO-LRFD-14.7.5.3.4 :

i) Shear Deformations due to Related Forces of Elastomeric Bearing under Service Limit State (WSD) :

a)

Required.

b)

c) 57.000 mm

d) 22.24 mm

ii) Computation of Shear Deformation of Elastomeric Bearing due to Thermal Movment Forces :

Fixe Against Shear Deformation sS-Fix-Dev. = 5.065 Mpa due to Applied Total Loads (DL + LL) under Service Limit State (WSD) & the Shape Factor

Si = 8.333 is 3.400%

Sεi

Total Applied Loads (DL + LL) under Service Limit State (WSD) = %ei/100

Applid Loads under Service Limit State (WSD); d = Sεihri ; dAllow

dDev.-S

Applied Total Loads (DL + LL) under Service Limit State (WSD) = hri*sS-Dev./Ec

dDev.-L

Applied Live Loads (LL) only under Service Limit State (WSD) = hri*sL-Dev./Ec

Allowable Compressive Deflection of Bearing due to Applied Total Loads (DL + LL) under Service Limit State (WSD).

Allowable Compressive Deflection of Bearing due to Applied Live Loads (DL + LL) only under Service Limit State (WSD).

The Horizontal Movement of Bridge Superstruce, DO, shall be taken as the Extreme Displacement caused by Creep,Shirnkage, Post-tensioning combined with Thermal Effects Computed in accordance with Article 3.12.2. For Max.

Shear Defotmation of the Bearing, under Service Limit State (WSD) - DS shall be taken = DO with Modifications as

Elastomeric Bearing should have to Satisfay the Provision of Equation 14.7.5.3.4-1; hrt ³ 2DS; where,

hrt is Total Thickness of Elastomeric Layers in Bearing in mm = HEla hrt.

DS is the Maximum Shear Deformation of the Elastomeric Bearing due to DS

Related Forces of under Service Limit State (WSD)

a) The Total Shear Deformation of Elastomeric Bearing due to Thermal Movment 11.990 mm

b) a 0.000011

c) L 24,400.000 mm

e) T 35

iii) Computation of Shrinkage Strain under AASHTO-LRFD-5.4.2.3-3:

a)

b) t 28 days

c) 0.100

d) 0.86

e) -0.005

f) The Total Shear Deformation of Elastomeric Bearing due to Shrinkage (132.20) mm

iv) Computation of Total Deformation due to Temperature & Shrinkage for One Bearing End :

a) Total Deformation due to Temperature & Shrinkage for One Direction for Total 11.985 mm

b) 5.992 mm

v) Computation of Deformation due to Horizontal Force Caused by Vehicle Braking (AASHTO-LRFD-3.6.4):

a) Horizontal Braking Force upon Bridge Deck shall be taken as the Greater value ofi) 25% of the Axle Weight of Design Truck or Design Tandem or;ii) 5% of of the Design Truck + Lane Load or 5% of Design Tandem + Lane Load.

b) Total Unfactored Axle Weight of Design Truck for an Interior Girder 650.000 kN

c) 162.500 kN

d) Total Unfactored Weight of Design Truck + Lane Load for an Interior Girde 1,115.000 kN

DThermal

Forces is Expressed by DThermal = 1.3aL( TMax.-Design - TMin-Design); where(Equation-3.12.2.2.2a-1)

a is Co-efficient of Thermal Expansion of Concrete in mm/mm/OC mm/mm/OC

L is Expansion Length in mm ( Span Length of Bridge) = SL-C/C

(TMax.-Design - TMin-Design) is the Difference of Temperatue of the Locality = T OC

For Moist Cured Concretes devoid of Shrinkage-prone Aggregates, the Strain due Shrinkage is Express by the

Equation esk = (-) kskh *(t/(35.00 + t)*0.51*10-3,where,

t is Drying Time in days after Casting for Curing Concrete = 28days.

ks is Size Factor Specified in Figure-5.4.2.3.3-2. For Ratio V/S = 150, the ks

value of ks = 0.10

kh is Humidity Factor Specified in Table 5.4.2.3.3-1. For 80% of Humidity & kh

the value of kh = 0.1

Thus Strain due Shrinkag esk = -kskh *(t/(35.00 + t)*0.51*10-3 eSk

DSk

= esk*SL-C/C

D(T+Sk)-Total

Bridge Length = DThermal + DSk

Deformation due to Temperature & Shrinkage For Each End =D(T+Sk)-Total/2 DT+Sk

WTruck

25% of Unfactored Axle Weight of Design Truck = WTruck*25% WTruck-25%

WTruck+Lane

e) 55.750 kN

f) Prevaliled Load for Braking Force 162.500 kN

g) 32.500 kN

h) 2,000.000 N/mm

i) 16.250 mm

j) 22.24 mm

k) Relation betweek Total Thickness of Elastomeric of Bearing & Total Shear Deformation hrt.>2DS OK

6 Checking for Combined Compression & Rotation under Provision of AASHTO-LRFD-14.7.5.3.5 :

i) Provisions & Requirmens for Combined Compression & Rotation of Elastomeric Bearing :

a) Checking for Combined Compression & Rotation of Elastomeric Bearing should be done under Service Limit State of

b) Rotations shall be taken as the Maximum Sum of the Effects of Initial Lack of Parallelism & Subsequent Girder End Rotation due to Imposed Loads & Movements.

c) Bearings shall be Designed so that Uplift does not occour under any Combination of Load & Corresponding Rotation .

d) In Rectangular Bearings Restriction of Uplift should be verified the Provisions of under Mentioned Equation;

e) In Rectangular Bearings Subject to Shear Deformation should Satisfay the Proviasins under Mentioned Equation;

e) In Rectangular Bearings Fixed against Shear Deformation should Satisfay the Proviasins under Mentioned Equation;

h) n 4.000 nos.(If Thickness of Exterior Elastomer Layer is more than One-half Thickness of

i) 12.000 mm

j) 5.786 Mpa

5% of of the Design Truck + Lane Load = WTruck+Lane*5% WTruck+Lane-5%

FBrak-Preva.

Braking Force Applicable for 1 (One) no. Girder = FBrak-Preva./NGir. FBrak.

Shear Stiffness of Elastomric Bearing, Kq = L*W*G/HEla Kq.

Deformation due to Shearing Force in One Direction = FBrak/Kq DBrak

Total Shear Deformation for All Applied Forces = DT+Sk + DBrak DS

Design (WSD).

sS > 1.00GS(qS/n)(B/hri)2; Equation-14.7.5.3.5-1.

sS < 1.875GS[(1- 0.20(qS/n)(B/hri)2]; Equation-14.7.5.3.5-2.

sS < 2.25GS[(1- 0.167(qS/n)(B/hri)2]; Equation-14.7.5.3.5-3. Where for all the Equations,

n is Interior Layers of Elastomer in Bearing = NEla-Inn

Interior Layer in that case the Parameter n should be replaced by (n + 1). For

the present case Thicness of Interior Layers, tEla-Inn = 12.00mm & Thickness

of Exterior Layers, tEla-Out =4.50mm. Since tEla-Out < 1/2tEla-Inn thus it requairsno change in Parameter n).

hri is Thickness of Each Internal Elastomeric Layers in mm = tEla-Inn hri.

sS is the Developed Stress in Elastomer in MPa due to Total Applied Loads sS

(DL + LL) under Service Limit State (WSD) = sS-Dev.

k) B 300.000 mm

l) 1.588E-15 Rad.

ii) Chicking Provisions of Up-lift Requirmens Elastomeric Bearing According to Equation-14.7.5.3.5-1.

a) 1.683E-12 Mpa(Equation-14.7.5.3.5-1).

b) sS>sS-Up-lift Ok

iii) Chicking Provisions of Shear Deformation of Elastomeric Bearing According to Equation-14.7.5.3.5-2.

a) 12.554 Mpa(Equation-14.7.5.3.5-2).

b) sS<sS-Shear-Def Ok

iv) Chicking Provisions for Elastomeric Bearing Fixed Against Shear Deformation under Equation-14.7.5.3.5-32.

a) 15.268 Mpa(Equation-14.7.5.3.5-2).

b) sS<sS-Fix-Shear Ok

7 Checking for Stablity of Elastomeric Bearing under Provision of AASHTO-LRFD-14.7.5.3.6 :

i) Provisions & Requirmens for Stablity of Elastomeric Bearing :

a)

b)Investigation in Respect of Stability is required.

c)

d)

ii) Provisions Stablity of Elastomeric Bearing According to Equation-14.7.5.3.6-1. :

a)

B is Length of Bearing Pad if Rota ion is about Bearing's Transverse Axis, orB is Width of Bearing Pad if Rota ion is about Bearing's Longitudinal Axis,

For the Present Case Rotation is Bearing's Longitudinal Axis & B = WBear.

qS is Maximum Rotation due to Total Applied Loads (DL + LL) under Service qS

Limit State (WSD) in Red. On Support Position of Girder = qTotal-WSD

Computed Stress for Up-lift = 1.00GS(qS/n)(B/hri)2; sS-Up-lift

Relation between Develop Stress, sS & Computed Stress for Up-lift sS-Up-lift

Shear Deformation Stress = 1.875GS[(1- 0.20(qS/n)(B/hri)2] sS-Shear-Def

Relation between Develop Stress, sS & Computed Shear Deformation sS-Shear-Defor

Fixed Shear Deformation Stress = 2.25GS[(1- 0.167(qS/n)(B/hri)2] sS-Fix-Shear.

Relation between Develop Stress, sS & Computed Shear Deformation sS-Shear-Defor

Bearing shall be investigated for Instability under Service Limit State (WSD) Load Combination.

Bearing should Satisfy the under Mentioned Equation-14.7.5.3.6-1. If the Provision of Equation Satisfy, than further

For a Rectangular Bearing having value of L is Greater than W, the Stability shall be Investigated by Interchanging their Position in Equation-14.7.5.3.6-2 & Equation-14.7.5.3.6-3.

For Circular Bearing the Stability should be Investigated by using the Equation-14.7.5.6-1/2/3 of Rectangular Bearingwith W = L = 0.80*D, where D is Diameter of Circular Bearing.

Euation of Stability is 2A £ B; in which

b) A 0.421

c) B 0.234

d) G 0.950 Mpa

e) L 300.000 mm

f) W 300.000 mm

g) 57.000 mm

h) 7.143

i) A>B Not SatifyRequires Further Investigation

j)

iii) Provisions Stablity of Elastomeric Bearing According to Equation-14.7.5.3.6-4. :

a)

b) GS/(2A-B) 11.145 Mpa

c) sS<GS/(2A-B) Satisfied

iv) Provisions Stablity of Elastomeric Bearing According to Equation-14.7.5.3.6-5. :

a)

b) GS/(A-B) 36.169 Mpa

c) sS<GS/(A-B) Satisfied

v) Since Provisions of Equation-14.7.5.3.6-4 & Equation-14.7.5.3.6-5. Are being Satsfied, thus the ElastomericBearing is Safe in Respect of Stability.

8 Checking for Reinforcement of Elastomeric Bearing under Provision of AASHTO-LRFD-14.7.5.3.7 :

i) Provisions & Requirmens for Reinforcement of Elastomeric Bearing :

a)

A = (1.92hrt/L)/(1+2.0L/W)1/2

B = 2.67/((S +2.0)(1 + L/4.0W)), where;

G is the Shear Modulus of the Elastomer in Mpa.

L is Length of Rectangular Elastomeric Bearing along Bridge Longitudinal

Axis in mm (Parallel to Traffic) = Blearing.

W is Width of Bearing in Transverse Direction in mm = WBearing.

hrt is Total Thickness of Elastomeric Layers in Bearing in mm = HEla hrt.

S is the Shape Factor for Elastomeric Bearing = Si S.

Relation between Calculated A & Calculated B according to Equattion Provision.

Since Provisions of Equation-14.7.5.3.6-1 have not Satisfied, thus it requires Investigations According to Provisions of Equation-14.7.5.3.6-4 & Equation-14.7.5.3.6-5.

If the Bridge Deck is Free to Translate Horizontally, than Provisions of under Mentioned Equartion should Satisfy;

sS £ GS/(2A-B) . (Equation-14.7.5.3.6-4)

Calculated Value of GS/(2A-B)

Relation between Develop Stress, sS & Computed value of GS/(2A-B)

If the Bridge Deck is Fixed to Translate Horizontally, than Provisions of under Mentioned Equartion should Satisfy;

sS £ GS/(A-B) . (Equation-14.7.5.3.6-5)

Calculated Value of GS/(A-B)

Relation between Develop Stress, sS & Computed value of GS/(2A-B)

The Thickness of the Steel Reinforcement, hs, shall have to Satisfy the Provisions of Equation-14.7.5.3.7-1 under

Service Limit State (WSD) as Mentioned here in, hs ³ 3hMax sS/Fy

b)

c) 3.000 mm

d) 165.000 Mpa

e) 12.000 mm

f) 2.150 Mpa

g) 5.786 Mpa

h) 410.00 Mpa

ii)

a) 0.508 mm

b) hs>3hmax sS/Fy Satisfied

iii)

a) 0.313 mm

b) hs>2.0hMax sL/DFTF Satisfied

iv) Since Provisions of Equation-14.7.5.3.7-1 & Equation-14.7.5.3.7-2. Are being Satsfied, thus the ElastomericBearing is Safe in Respect of Reionforcement as Steel Laminate.

The Thickness of the Steel Reinforcement, hs, shall have to Satisfy the Provisions of Equation-14.7.5.3.7-2 under

Fatigue Limit State as Mentioned here in, hs ³ 2.0hMax sL/DFTH, where;

hS is Thickness of Steel Laminate in Steel-Laminated Elastomeric Bearing in hS

mm = tS-Inn

DFHT is Constant Ampletude Fatigue Threshold for Category-A as Specified DFHT in Article-6.6 im MPa. According to Article-6.6.1.2.5, In Provision of Fatigue Resistance & Respective Table--6.6.1.2.5-3 unser Catgory-A, the value of

Constant Ampletude Fatigue Threshold, DFHT = 165.00 Mpa.

hMax is the Thickness (mm) of Thickest Elastomeric Layer in Elastomeric hMax.

Bearing .Here Internal Elastomeric Layers are Thickest with Equal = tEla-Inn

sL is Develop Average Stress in Mpa on Elastomeric Bearing Under Service sL

Limit State (WSD) only due to Live Loads (LL) = sL-Dev.

sS is Develop Average Stress in Mpa on Elastomeric Bearing Under Service sS

Limit State (WSD) for Total Applied Loads (DL + LL) = sS-Dev.

Fy is Yield Strength of Steel as Reinforcement of Elastomeric Bearing with a Fy

value = fy Steel Ultimate Strength (60 Grade Steel).

Computed values of Equation-14.7.5.3.7-1, hs ³ 3hMax sS/Fy

Calculated value of 3hMax sS/Fy 3hMax sS/Fy

Relation between hS, the Thickness of Steel Laminate in Steel-Laminated

Elastomeric Bearing & Calculated value of 3hMax sS/Fy

Computed values of Equation-14.7.5.3.7-2, hs ³ 2.0hMax sL/DFTH

Calculated value of 2.0hMax sL/DFTH 2.0hMax sL/DFTF

Relation between hS, the Thickness of Steel Laminate in Steel-Laminated

Elastomeric Bearing & Calculated value of 2.0hMax sL/DFTF

All Critical Load Combinations should considered for Designing of Bearings in Respect of Translational & Rotational .

Calculated Maximum Rotation for Bearing shall be the Total Rotation due to Unfactored Service Loads. This will have

R Design of RCC Bearing Pads for Abutment :



i)

9.807






ii)






iii) Design Data for Resistance Factors for Conventional Construction (AASHTO LRFD-5.5.4.2.1). :










j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85


a) 21.000 MPa

b) 8.400 MPa



gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.




c


c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa

v) Strength Data related to Working Stress Design & Service Load Condition ( WSD & AASHTO-SLS ) :

a) 8.384 n 8

b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102

2 Dimentional Data of RCC Bearing Pads for Abutment :

i) Provided Dimensions of RCC Bearing Pad for Central Girder :

a) Length of Bearing Pad (Parallel to Traffic) 500 mm

b) Width of Bearing Pad (Transverse to Traffic) 930 mm

c) Depth of Bearing Pad (Under Central Girder) 115 mm

ii) Provided Dimensions of Elastomeric Bearing Pad for Central Girder :

a) Length of Bearing Plate (Parallel to Traffic) 400 mm

b) Width of Bearing Plate (Transverse to Traffic) 300 mm

c) Depth of Bearing Plate 72 mm

d) Clear Cover on Vertical Faces 25 mm

e) Clear Cover on Horizontal Face on Top 25 mm

f) Width of Abutment Cap 1000 mm

g) Depth of Abutment Cap 600 mm

3 Philosophy in Structural Design of RCC Bearing Pad :

a) The RCC Bearing Pads for Girders are absolutely Compressive Components & their Structural shall be accordingly.


c Ec











Value of R = 0.5*(fckj) = 0.5*(8.400*0.307*0.898) = 1.156

LPad

WPad

hPad

LB-Plate

WB-Plate

hB-Plate

CCov.-V

CCov.-H

bAb-Cap

hAb-Cap

b)

c)

d) Since Total Vertical Loads from Girder would be Placed Directly upon the Elastomeric Bearing Pad, afterward uponthe RCC Pad & Finally to the Abutment Cap. Thus it is require to Design Flexural Design as Compression Member for Reinforcement & Subsequent Checking for Shear.

e)

4 Applied Factored Loads upon Bearing Pad for Central Girder :

a) 1,149.477 kN

b) 694.340 kN

5 Structural Design of RCC Bearing Pad for Central Girder under Strength Limit State (USD) :

i) Structural Design of Compression Reinforcement as Short Column :

a) 120000.000

b) 1,149.477 kN

1149.477*10^3 N

c) (2,420.787)

d) The (-) ve value of As-req indicates that No Compression Reinforcement is require for RCC Pad. But for Safe Guardof RCC Pads the under Mentioned Reinforcements are being Proposed to Provide.

e) 12 mm

f) 113.097

g) Let Provide the Vertical Bars on Width Faces (Transverse to Traffic) 100 mm C/Cof RCC Pad having Spacing 100 mm C/C.

h) Number of Vertical Bars reqired on Width Face of RCC Pad 9.800 nos

i) Let Provide 8 nos. Vertical Bars on Each Width Face of RCC Pad 10 nos.

j) Let Provide 2 (Two) Rows of Vertical Bars on Length Faces 100 mm C/C(Parallel to Traffic) of RCC Pad having Spacing 100 mm C/C.

Since the Pads are of very short Depth, thus the Structural Design will based on Short Column Phenomenon.

Design shall be according to Provisions of AASHTO-LRFD-5.7.5 & the Design Phenomenon for RCC Structure.

Initial Structural Design will be done under Strength Limit State (USD) & Subsequent Checking will done based onService Limit State (WSD).

Applied Factored Loads (DL + LL) under Strength Limit State (USD) PUSD

Applied Factored Loads (DL + LL) under Service Limit State (WSD) PWSD

Surface Area of RCC Pad under Elastomeric Bearing Pad = LB-Plate *WB-Plate AC mm2

Vertical Concentrated Load upon the Elastomeric Bearing Pad PUSD is the PU

Ultimate Load for Design. Thus PUSD = PU

Steel Area Required for Compressive Member = (PU - 0.85f/cAC)/fy AS-req mm2

Let Provide 12f Bars as Reinforcement for RCC Pad on all Faces DBar

Cross-sional Ares of 12f Bar = pDBar-Comp2/4 Af-12 mm2

sW-Pad-pro-V

NBar-W-req

= (WPad -2*CCov-V)/sW-Pad-pro-V + 1

NBar-W-V-pro

sL-Pad-pro-V

k) Number of Compression Bars reqired on Length Face of RCC Pad 4.500 nos

l) Let Provide 5 nos. Bars on Length Face of RCC Pad for Each Row 5 nos.

ii) Arrangement of Tie Reinforcement :

a) 10 mm

b)Diameter or 48 times of Tie Bar Diameter or the Least Dimension of Compressive Member.

c) 192 mm

d) 480 mm

e) 500 mm

f) 100 mm C/C

iii) Arrangement of Reinforcement on Top Surface of of RCC Bearing Pad :

a) Alternate Compression Bars from Inner Row of Length Faces would be bend Horizontally and will Continue up to the Outer Row on Opposite Face.

b) Each Compression Bar of Outer Row on Length Faces would Continue up to Guide Wall Top & then after those will Turned Downward to Continue through Inner Row Position to Enter inside of the Horizontal Face of RCC Pad.

c) Each Compression Bar on Width Face will Bend Horizontally to Continue up to opposite Face and again those willBend Downward to become the Compression Bar for that Face.

6 Structural Design of RCC Bearing Pad for Central Girder under Service Limit State (WSD) :

a)

b) P 694.340 kN

c) 8.400 MPa

d) 164.000 MPa

e) 120000.000

f) (1,912.56)

NBar-L-req

on Eacg Row = (LPad -2*CCov-V)/sL-Pad-pro-V+ 1

NBar-L-Face-1Set-pro

According to ACI Code against 12f Compression Bars the Tie Reinforcement DBar-Tie

will of 10f Bars.

Under same Code. Spacing of Tie Reinforcement with of 10f Bars will be the Lesser value of 16 times of Main Bar

16 times of Main Bar Diameter = 16*DBar-Comp sMain

48 times of Tie Bar Diameter = 48*DBar-Tie sTie

Least Dimension of RCC Pad = LPad sDimn

Let Provide 100 mm C/C Spacing for 10f Tie Bars . sTie-pro

Under Service Limit State or WSD the Applied Load upon Bearing Pad is Expressed by P = fc Ac + fsAs; where,

P is th Applied Concentrated load upon Bearing Pad = PWSD

fc is Concrete Allowable Strength under Service Limit State (WSD) = 0.40f/c fc

fs is Steel Allowable Strength under Service Limit State (WSD) = 0.40fy fs

Ac is Area of RCC Pad under Elastomeric Bearing Pad = LB-Plate *WB-Plate AC mm2

As is Total Cpompresive Steel Area Required for RCC Pad against Applied AS mm2

Load = (P - Acfc)/fs

g) The (-) ve value of As-req indicates that No Compression Reinforcement is require for RCC Pad. But for Safety of

5 Checking for Shear of RCC Bearing Pad according to Article AASHTO-LRFD-5.7.5 :

i) Provisions of Article AASHTO-LRFD-5.7.5 :

a) Sketch Diagram of RCC Bearing Pad:-Elastomeric Bearing Pad

150 mm 150 mm Girder

115mm 530 mm

930 mm930 mm

b) 3,855.600 kN 3855.600*10^3 N

c) 4,284.000 kN(Equ.-5.7.5-2). 4284.000*10^3 N

d) f 0.900

e) 120000.000

f) m 2.000 i) If the Supporting Surface is Wider than the Loaded or Bearing Device Area

m-1 2.784 >2.00

ii) If the loaded Area is subject to Non-uniformly Distributed Load in that Case m-2 NA

g) 930000.000

h) Pu<Pr

SatisfiedSatisfied or Not.

i)

Pads Reinforcements are being Proposed under to Provide as mentioned on Strength Limit State (USD) Section.

W =

The Factored Bearing Resistance of Bearing Device Pr = fPn; in which, Pr

(Equ.-5.7.5-1).

Pn = 0.85f/cA1m is Nominal Bearing Resistance in N, Pn

f is Multipluing Factor for Shear

A1 is Area under Bearing Device in mm2 = AC A1 mm2

m is a Modification Factor related to Area A2

m = Ö(A2/A1) £ 2.00 (The Present Case Wider Loaded Area with Uniformly Distributed Load).(Equ.-5.7.5-3).

m = 0.75 Ö(A2/A1) £ 1.50 (Equ.-5.7.5-4).

A2 is Notational Area in mm2 within RCC Pad & Abutment = W*bAb-Cap A2 mm2

Relation between Applied Factored (USD) Load upon Bearing Pad PU & Computed

Factored Bearing Resistance Pr.& whether the Provisions of Equation.-5.7.5-3 have

Since all required Provisions are being Satisfied, thus the Design of RCC Pad is OK.

S. Calculation of Deformations (Deflections & Rotations) of Different Structural Elements due to Applied Load, Shrinkage and Temperature :


1 Structural Data :

































ii) Number of Traffic Lane on Bridge Deck & Flange Width :

a) 2.028 nos

SL

SAddl.

LGir.

WCarr-Way.

WS-Walk.

WCurb.

WR-Post.

WB-Deck.

RW&D.

PW&B.

hR-Post.

hCurb.

Rnos.

C/CD-R-Post.

tSlab.

tWC

NGirder.

NX-Girder.

hGir.

hX-Girder.

b-Gir-Web.

bX-Gir-Web.

C/CD-Girder.

C/CD-X-Girder.

CD-Ext.-Girder-Edg.

ClD-Int.-Girder.

FM-Girder-V.

FM-Girder-H.

FX-Girder-V.

FX-Girder-H.

ASup-Vert. m2



b) 2.000 m

2 Design Data :





















3 General Data for Construction Materials of Different Structural Components :

i)

9.807


Calculated Flange Wideth of Interior T-Girder = bFln bFln.

DAxel.

DWheel.

LLRW-Load

LLRS-Load

LLMW-Load

LLMS-Load

LLFW-Load

LLFS-Load

LLLane

LLLane-Int.

LL-Pedest N/mm2









ii)







a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa


a) n 8

b) r 19.524c) k 0.291 d) j 0.903

e) R 1.102






e)

gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


c

Concrete Allowable Strength under Service Limit Staete of Design (WSD) = 0.40f/c fc


c Ec







Steel Allowable Strength under Service Limit Staete of Design (WSD) = 0.40fy fs





(Respective Resistance Factors are mentioned as f )

fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

Flexural value of f of Compression Member will Increase Linearly as the Factored Axil Load Resitance,

f)

g) For Bearing on Concrete 0.700

h) For Compression in Strut-and-Tie Modeis 0.700

i) For Compression in Anchorage Zones with Normal Concrete 0.800

j) For Tension in Steel in Anchorage Zones 1.000

k) For resistance during Pile Driving 1.000

l) For Partially Prestressed Components in Flexural with or without Tension 1.000 m)

n) PPR

o)

p)

q) 410.000

r)

vi)

a) 0.85

Acc =Area of Concrete Element in Compression of Crresponding Strength.

b) 0.85



a)

b)

c)Here:

d)e)

f)g) For Strength Limit State;

h) 1.000

i) 1.000

j) 1.000

k)

l)

m)

fPn, Decreases from 0.10f/cAg to 0.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.

fFlx-PPR.

Resistance Factor f = 0.90 + 0.10*(PPR) in which,

PPR = Apsfpy/(Apsfpy + Asfy), where; PPR is Partial Prestress Retio.

As = Steel Area of Nonprestressing Tinsion Reinforcement in mm2 As mm2

Aps = Steel Area of Prestressing Steel mm2 Aps mm2

fy = Yeiled Strength of Nonprestressing Bar in MPa. fy N/mm2

fpy = Yeiled Strength of Prestressing Steel in MPa. fpy N/mm2

b Factors for Conventional RCC & Prestressed Concrete Design (AASHTO LRFD-5.7.2.2). :

Flexural value of b1, the Factor of Compression Block in Reinforced Concrete b1

up to 28 MPa.i) For Further increases of Strength of Concrete after 28 MPa agaunst each 7MPa

the value of b1 will decrese by 0.05 & the Minimum Value of b1 will be 0.65.

ii) For Composite Concrete Structure, b1avg = Σ(f/cAccb1)/Σ(f/

cAcc); where,











Qi = Force Effect,



ii) Permanent & Dead Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1 ; Table 3.4.1-1&2 :








iii) Live Load Multiplier Factors for Strength Limit State Design (USD) According to AASHTO-LRFD-3.4.1; Table 3.4.1-1&2 :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750

i) 1.000




gEH


gEV


gES












k) 1.000






q) -

r) -

t) 1.000





















gEH


gEV


gES





b) 1.000


d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






















q) -

r) -

t) 1.000

6 Provisions of AASHTO-LRFD-5.7.3.6.2 in Flexural Deformation of Components for Deflection & Camber, :

i) Criterion for Deflection :

a)the followings;

i) Consideration of Deflection is Mandatory for Provision of Orthotropic Decks;

ii) Consideration of Deflection is Mandatory for Provision of Precast Reinforced Concrete Three-side Structure According

iii) Consideration of Deflection is Mandatory for Provision of Metal Grid Decks and other Light Weight Metal and Concrete

b) In Computation of Deflection the Service Limit State should be Considered in Calculation of Moments against Respective

c)

d) According to AASHTO-LRFD-2.5.2.6.2 the Deflection Limit for Steel, Aluminum and/or Concrete Constructions with different Combination of Loads are the Followings :

L/800 30.500 mm

L/1000 24.400 mm

L/300 81.333 mm

L/1000 65.067 mm

ii) Axil Deformation of Components, (AASHTO-LRFD-5.7.3.6.3) :

a) Instantaneous Deformation by Shortening or Expansion due to Loads shall be determined by using the Modulus of

b) Instantaneous Deformation by Shortening or Expansion due to Temperature shall be determined in accordance with

c) Long-term Deformation by Shortening due to Shrinkage & Creep shall be determined in accordance with Provisions




According to AASHTO-LRFD-2.5.2.6.2 Consideration of Deflection is Optional Item for RCC Bridge Structure Except for

to Article 12.14.5.9;

Bridge Deck subject to the Serviceablity Provisions of Article 9.5.2.

Applied Loads. The Dynamic Load Allowance IM should be Considered for Respective Live Loads in Combination with theService-I.

Provisions of Article -2.5.2.6.2 should be Considered for Analysis of Respective Deflections.

i) For general Vehicle Load, D = L (Span Length)/800

ii) For Vehicle and/or Padestrain Loads, D = L (Span Length)/1000

iii) For Vehicle Load on Cantilever, D= L (Span Length)/300

iv) For Vehicle and/or Padestrain Loads on Cantilever, D = L (Span Length)/375

of Elasticity of Materials (Concrete-Ec & Steel-Es) at the Time of Loading.

Provisions of AASHTO-LRFD-3.12.2; 3.12.3 & 5.4.2.2.

of AASHTO-LRFD- 5.4.2.3.

iii) Provisions for Instantinous Deflection at Mid Span :

a)

b)

c)

d)

e)in which;

f)

g)

h) N-mm

i)

j) mm

k) N-mm

l)it will be at Support Position for the case of Cantilever Spans.

m)(+) ve. & (-) ve. Moments respectively.

iv) Provision Long-time Deflection at Mid Span (AASHTO-LRFD-5.7.3.6.2) :

a)

b) l

c) 4.000

d) 3.000

Deformation/ Deflection of Structural Components in respect of Dead & Live Load, Creep, Shrinkage, Thermal Changes, Settlement & Prestressing should be considered according to provision of AASHTO-LRFD-2.5.2.6.

In Calculation of Deflection & Chamber Dead Load, Live Load, Prestressing, Erection Load, Concrete Creep & Shrinkage and Steel Relaxation should be Considered. For Determining Deflection & Camber the Provisionsof AASHTO-LRFD-4.5.2.1; 4.5.2.2 & 5.9.5.5. are applicable.

Computation of Instantaneous Deflection should be based on Modulus of Elasticity of Conncrete-Ec as mentioned

in AASHTO-LRFD-5.4.2.4. & Flexural Regidity of the Component EcI. For the purpose either the Gross Moment of

Inertia-Ig or the Effective Moment of Inertia-Ie of the Component should ce considered. Here;

EcI is Flexural Regidity of the Component as mentioned EcI

The Effective Moment of Inertia, Ie = (Mcr/Ma)3Ig + (1 - (Mcr/Ma)3)Icr £ Ig Ie mm4

Ig is the Gross Moment of Inertia for the Section under consideration in mm4 Ig mm4

Icr is the Moment of Inertia of the Crack Section under consideration in mm4 Icr mm4

Mcr is Crack Moment for the Section having value = fr(Ig/yt) Mcr

fr is Modulus of Rupture of Concrte as per AASHTO-LRFD-5.4.2.6 in Mpa fr N/mm2

yt is the distance from the Neutral Axie to the Extrim Tension Fiber in mm yt

Ma is the Max. Moment of Component for which the Deformation Computed Ma

In Prismatic Member the Effective Moment of Inertia- Ie will be at Middle of Simple or Continuous Spans, whereas

In Nonprismatic Member the Effective Moment of Inertia- Ie will be the Average of values at Critical Sections due to

The Long-time Deflection is Expressed by DLong = l*DInst. In which; DLong.

l is a Multiplying Factor for the Respective Cause of Deflection

If Instantaneous Deflection is Based on Ig, then Facto l = 4.00 l-Ig.

If Instantaneous Deflection is Based on Ie, then l-Ie.

Facto l = 3.00 - 1.2(A/s/As) ³ 1.60. Where,

e) 0.000

f) 14,476.459

v) Axil Deformation of Components, (AASHTO-LRFD-5.7.3.6.3) :

a) Instantaneous Deformation by Shortening or Expansion due to Loads shall be determined by using the Modulus of

b) Instantaneous Deformation by Shortening or Expansion due to Temperature shall be determined in accordance with

c) Long-term Deformation by Shortening due to Shrinkage & Creep shall be determined in accordance with Provisions

vi) Provisions for Instantaneous Deflection at Mid Span due to Application of Different Loads (DL & LL) :(Provisions of "Deformation of Concrete Structure" by Dan Earle Branson; Chapter-3-1, Deflection of Concrete Beams And One-Way Slab, And Both Steel And Reinforced Concrete Composite Beams. Also Provisions of Strength of Materials" by ARTHUR MORLEY, O.B.E. Chapter-3-1, Deflection of Beams)

a) mmcaused under Service Load Condition due to Individual Application of Dead

b)

(Provisions of "Deformation of Concrete Structure" by Dan Earle Branson: Chapter-3-1 Deflection of Concrete Beams And One-Way Slab, And Both Steel And Reinforced Concrete Composite Beams. Equation-3-1)

c)

(Provisions of Strength of Materials" by ARTHUR MORLEY, O.B.E. Chapter-3-1, Deflection of Beams)

c-i) 0.104

c-ii) 0.104

c-iii)Reference Support End & Position of Deflection Point (Mid Span), the value

c-iv)

A/s = Area of Compression Reinforcement in mm2. For Simple Supported Singly A/

s.

Reinforcer T-Girder; A/ = 0

As = Area of Nonprestressed Tension Reinforcement in mm2 As. mm2

of Elasticity of Materials (Concrete-Ec & Steel-Es) at the Time of Loading.

Provisions of AASHTO-LRFD-3.12.2; 3.12.3 & 5.4.2.2.

of AASHTO-LRFD- 5.4.2.3.

Instantaneous Deflection is Expressed by DInst. = Summation of Deflections DInst.-Total

Loads (DL) & Live Loads (LL). The Maximum Deflection due to simultaneousapplication of all Loads (DL & LL) occurs at Mid of Simple Supported Span.

The Instantaneous Defection on Simple Supported T-Girder Bridge on its Interior Beam at Mid Span Caused by Applied Uniformly Distributed Loads (UDL) or Concentrated Loads (CL) due to Dead Loads (DL) & Live Loads (LL) is Expressed

by the Equation, DInst-UDL&CL= KMaL2/EcIe. Here,

K is Deflection Coefficient which depends upon the Nature of Applied Load (DL & LL) whether Uniformly Distributed Load, (UDL) or Concentrated Load (CL). In case of Concentrated Load (CL) Deflection Coefficient also depends upon Positionof Load (DL & LL) in respect of Point of Deflection from One Particular Reference Support End.

For Uniformly Distributed Loads (DL & LL) value of K = 5/48 KUDL-DL&LL

For Concentrated Loads (DL & LL) Applied at Mid Span value of K = 5/48 KCL-DL&LL

If Concentrated Load (DL & LL) is Applied at a Distance in between KCL-DL&LL(X<L/2)

of K is Expressed by K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24XL2); Here, X is Distance of Applied Concentrated Load (DL & LL), L is Span Length. The Provision is Applicable for the Case when X < L/2.

If Concentrated Load (DL & LL) is Applied at a Distance beyond the KCL-DL&LL(X>L/2)

Position of Deflection Point (Mid Span) from the Reference Support End,

d) L 24.400 m

e) N-mm

f) Modulus of Elasticity of Concrete, 23,855.620 MPa

g) 0.269

7 Calculation of Factored Dead Loads (DL) & Live Loads (LL) on Intermediate Girder under Service Limit State :

i) Sketch Diagram Showing Different Loads (DL & LL) on Intermediate Girder Under Service Limit State:


6.100 m 6.100 m 6.100 m 6.100 m

12.200 m0.300 6.100 m 0.300

m 108.750 kN c.g. of Wheels. m CL of Bearing 0.728 108.750 kN CL of Bearing

12.200 m8.628 m

12.928 m

17.228 m 26.250 kNm

A B

28.710 kN/m

6.200 kN/m

24.400 m

25.000 m


a) 25.260 kN/m Self Wt. & Attachments (Without WC & Utilies) for per meter Length of Girder.

the value of K is Expressed by K = (L/8 - (L - X)2/6L2; Here,

the value of K is Expressed by K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24(L-X)L2);Here, X is Distance of Applied Concentrated Load (DL & LL), L is Span Length. The Provision is Applicable for the Case when X > L/2.

L is the Span Length (C/C Distance between Bearings/Supports) = SL

Ma is Moment for the respective Load under Service Limit Staete of Design (WSD) Ma

Ec

Ie is the Effective Moment of Inertia of Girder Crack Section under Service Limit Ie mm4

State of Design (WSD).

X-Gir-3 =

X-Gir-2 =

L /2 = Rear Wheel under c.g Provisions.X-Rear = Midd. Wheel under c.g Provisions.X-Mid. = Front Wheel under c.g Provisions.X-Front =

å FDLint-All =

CL of Girder å FLLint-Lane-Ped =

LSpan =

LTotal =

Uniformly Distributed Dead Load (DL) on.Interior Girder due to DLIntt-Gir-Self

å FDLInt = 28.7100kN/m

å FLLInt. = 6.200kN/m

b) 3.450 kN/m meter Length of Girder


d) Sumation of Uniformly Distributed Dead Loads on Interior Girder (a + b) for 28.710 kN/m

iii) Live Loads (LL) on 1 no. Interior Girder due to Wheel Load & Lane Load according to Provisions of AASHTO-LRFD-3.6.1.2.2, 3.6.1.2.4 & 3.6.1.6 :

a) 26.250 kN

b) 108.750 kN

c) 108.750 kN

d) 6.200 kN/m

iv) Factored Dead Loads on 1 no. Interior Girder from Different Components & Attachments :

a) 25.260 kN/m

b) 3.450 kN/m

c) 17.053 kN

d) 28.710 kN/m Girder (a + b) for per Meter Length of Girder.

v) Factored Live Loads of Different Components for 1 no. Interior Girder :

a) 26.250 kN

b) 108.750 kN

c) 108.750 kN

d) 6.200 kN/m

Uniformly Distributed Dead Load (DL)on Interior Girder from WC. for per DLInt-Gir-WC.

DLInt-X-Gir.

åDL-Int.UD

Concentrated Live Load (LL) on 1 no.Interior Girder from Front Wheel LLInt-Wheel-Front.

Concentrated Live Load (LL) on 1 no.Interior Girder from Mid. Wheel LLInt-Wheel-Mid.

Concentrated Live Load (LL) on 1 no.Interior Girder from Rear Wheel LLInt-Wheel-Rear.

Uniformly Distributed Live Lane Load (LL) on 1 no. Interior Girder LLInt-Lane.

Factored Dead Load (DL)on Interior Girder due to Self Wt.& FDLIntt-Gir-Self

Attachments (Without WC) for per Meter Length = gDC*DLInt-Gir-Self

Factored Dead Load (DL) on Interior Girder due to WC. & Utilities FDLInt-Gir-WC+ Utility.

for per Merter Length = gDW*DLInt.-Gir-WC

Factored Concentrated Dead Load (DL) on Interior Girder from to 1 no. FDLInt-X-Gir.

Cross Girder = gDC*DLInt-X-Gir.

Sumation of Factored Uniformly Distributed Dead Loads (DL) on Interior åFDL-Int.UD-DL

Factored Concentrated Live Load (LL) on 1 no.Interior Girder from FLLInt-Wheel-Front.

Front Wheel = mgLL-Truck*LLInt-Wheel-Front

Factored Concentrated Live Load (LL) on 1 no.Interior Girder from FLLInt-Wheel-Mid.

Middle Wheel = mgLL-Truck*IM*LLInt-Wheel-Mid

Factored Concentrated Live Load (LL) on 1 no.Interior Girder from FLLInt-Wheel-Rear

Rear Wheel = mgLL-Truck*IM*LLInt-Wheel-Rear

Factored Uniformly Distributed Live Lane Load (LL) on 1 no. Interior Girder FLLInt-Lane.

= mgLL-Lane*LLInt-Lane

8 Calculation of Moments due to Factored Loads (DL & LL) at Mid Span of Intermediate Girder :

i) Moment due to Uniformly Distributed Loads (DL & LL) on Interior Girder :

a) 2,136.598 kN-m

b) 461.404 kN-m

ii) Moment due to Concentrated Loads (DL & LL) on Interior Girder :

a) 0.000 kN-m

b) 52.011 kN-m

c) 52.011 kN-m

d) 94.137 kN-m

e) 702.943 kN-m

f) 469.131 kN-m

9 Values of Deflection Coefficient 'K' for Different Applied Loads to Compute the respective Deflections at Mid Span of Intermediate Girder :

a) 0.104 (Applicable for Dead Loads of Main Girder, Slab, W-Course & Lane Live Load).

b) 0.000Girders having Position at Support Position the value = 0.

c) 0.475 Cross Girders having Position in between Referance Support & Mid of Span

d) 0.104 Girder having Position at Mid Span, the value = 5/48

Moments due to Uniformely Distributed Factored Dead Loads (DL) MUD-DL.

= åFDL-Int.UD-DL*L2/8

Moments due to Uniformely Distributed Factored Live Lane Loads (LL) MUD-LL.

= FLLInt.Lane*L2/8

Moments due to Factored Concentrated Dead Loads (DL) for 1st & 5th MCDL-X-1&5.

Cross Girder = FDLInt.X-Gir.*L/2 - FDLInt.X-Gir.*L/2

Moments due to Factored Concentrated Dead Loads (DL) for 2nd & 4th MCDL-X-2&4.

Cross Girder (Case for X < L/2) = FDLInt.X-Gir.*X-Gir-2 /2

Moments due to Factored Concentrated Dead Loads (DL) for 3rd MCDL-X-3.

Cross Girder (Case for X = L/2) = FDLInt.X-Gir.*X-Gir-3 /4

Moments due to Factored Concentrated Wheel Live Load (LL) for MCLL-W-Front.

Front Wheel (Case for X > L/2) = FLLInt-Wheel-Front.*(L - X-Front)/2

Moments due to Factored Concentrated Wheel Live Load (LL) for MCLL-W-Mid.

Mid Wheel (Case for X > L/2) = FLLInt-Wheel-Mid.*(X-Mid)/2

Moments due to Factored Concentrated Wheel Live Load (LL) for MCLL-W-Rear.

Rear Wheel (Case for X < L/2) = FLLInt-Wheel-Rear.*(X-Rear)/2

Deflection Coefficient for Uniformly Distributed Load (DL & LL), the value = 5/48 KUD-DL&LL

Deflection Coefficient for Concentrated Load (DL) for 1st & 5th Cross KCL-DL-1&5-X-Gir-Supp

Deflection Coefficient for Concentrated Dead Load (DL) for 2nd & 4th KCL-DL2&4-X-Gir(X<L/2)

for X < L/2, the value of K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24XL2).

Deflection Coefficient for Concentrated Dead Load (DL) for 3rd Cross KCL-DL-3-X-Gir-Mid

e) 0.101

f) 0.088 Wheel having Position in between Referance Support & Mid of Span

g) 0.148 Wheel having Position in between Referance Support & Mid of Span

10 Moment of Inertia of Different Component Sections Related to Flexural Design under AASHTO-LRFD :

i) Events related to Flexural Design of Interior Girder against Max. Moment under Service Limit State :

a) Provided Steel Area at Mid Span against Max. (+) ve. Moment 14,476.459

b) Effective Depth of Provided Tensial Reinforcement 1,808.222 mm

c) 525.573 mm

d) 1,282.649 mm

e) Diameter of Provided Main Tensial Reinforcement Bars 32.000 mm

f) Number of Provided Main Tensial Reinforcement Bars 18.000 nos.

g) Number of Layers for Provided Main Tensial Reinforcement Bars 5.000 nos.

h) 160.000 mm 0.160 m

h) Transformed Steel Area as Equivalent Concrete Area Provided for Tensile 115,811.672

0.115811672

i) 723.823 mm 0.724 m

ii) Calculations for Moment of Inertia of Main T-Girder Section about its Natural Axis for Un-crack Section :

a)

2.000 m

0.200 m 0.712 m

Deflection Coefficient for Concentrated Live Load (LL) for Front KCL-LL-Wheel-Front(X>L/2)

Wheel having Position beyond Mid of Span for X > L/2, the value of

K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24(L-X)L2);

Deflection Coefficient for Concentrated Live Load (LL) for Middle KCL-LL-Wheel-Mid-(X<L/2)

for X > L/2, the value of K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24(L-X)L2).

Deflection Coefficient for Concentrated Live Load (LL) for Rear KCL-LL-Wheel-Rear-(X<L/2)

for X < L/2, the value of K = ((L-X)*(3L2 - 4(L - X)2) + (L - 2X)3)/(24XL2).

As-pro. mm2

dpro.

Depth of Neutral Axis from Top of Extreme Compression Face = kdpro dComp.

Depth of Tensial Bar Centriod from Neutral Axis = dpro - kd dTen

DBar

NBar

NLayer

Total Height of 5 Layers 32f Steel Bars if placed without Spacing = NLayer*DBar hSteel

As-Trans. mm2

Reinforcement = nAs-pro m2

Width of Transformed Steel Area as Equivalent Concrete Area = As-Trans./hSteel bSteel

Sketch Diagram of T-Girder Section :

bFln =

hFln = dNeut-Axis

0.612 m

0.900 m 2.000 m

0.388 m

0.350 m

b) 2.000 m

c) 0.200 m

d) 2.000 m

e) 0.350 m

f) 0.712 m

g) 0.612 m

h) 0.388 m

i) Moment of Inertia of Flange Portion about its Neutral Axis, 0.001

j) Moment of Inertia of Web Portion about its Neutral Axis, 0.170

k) 0.151

l) 0.265

m) 0.416

n) 0.416

4.161E+11

dFln =

hWeb/2 = hGir =

dWeb =

bWeb =

Flange Width of T-Girder bFln.

Flange Height of T-Girder. hFln.

Depth of T-Girder. hGir.

Width of T-Girder Web bWeb.

Distance of T-Girder Neutral Axis from Flange Top, dNeut-Gir. dNeut-Gir

= (bFal*hFln.*hFln./2+bWeb.*(hGir.-hFln.)*((hGir-hFln.)/2+hFln))/

(bFln.*hFln.+bWeb*(hGir-hFln.))

Distance between Neutral Axis of T-Girder & Neutral Axis of Flange Portion dFln.

= dNeut-Axis - hFln./2

Distance between Neutral Axis of Web Portion & Neutral Axis of T-Girder dWeb.

= ((hGir.- hFln.)/2+ hFln.)-dNeut-Axis

IFln. m4

IFln. = bFln.*hFln.3/12

IWeb. m4

IWeb. = bWeb.*(hWeb - hFln.)3/12

Moment of Inertia of Flange Portion about T-Girder Neutral Axis, IFln.-Neut. m4

IFln.-Neut = IFln. + bFln.*hFln.*dFln.2

Moment of Inertia of Web Portion about T-Girder Neutral Axis, IWeb.-Neut. m4

IWeb.-Neut = IWeb. + bWeb.*(hWeb-hFln.)*dWeb.2

Moment of Inertia of T-Girder about its Neutral Axis, = IFln.-Neut. + IWeb.-Neut. IT-Gir. m4

Gross Moment of Inertia for (Un-cracke) under Service Limit State = IT-Gir. Ig. m4

mm4

iii) Calculations for Moment of Inertia of T-Girder Crack Section about Natural Axis of Section :

a)

2.000 m

0.200 m

0.326 m 0.426 0.526 m

1.808 m 1.283 m 1.474 m

2.000 m

0.116

0.350 m

b) 2.000 m

c) 0.200 m

d) Depth of T-Girder Crack Section (Distance from Extreme Compression Fiber up kd 0.526 mto Flexural Neutral Axis of the Section).

e) 0.350 m

f) Distance between Neutral Axis of Crack Section & Neutral Axis of Flange Portion 0.426 m

g) 0.326 m

h) Distance between Neutral Axis of Crack Section & Centroid of Tensile 1.283 m


j) Moment of Inertia of Uncrack Web Portion about Neutral Axis of Crack Section 0.004

k) 0.074

Sketch Diagram of T-Girder Crack Section :

bFln =

hFln =

dWeb = dFln. = kd =

dpro = dTen = yt =

hGir =

As-Trans. = m2

bWeb =

Flange Width of T-Girder bFln.

Flange Height of T-Girder. hFln.

Width of T-Girder Web bWeb.

dFln.

= kd - hFln./2

Depth of Un-crack Web Portion of T-Girder dWeb.

= (kd.- hFln.)

dTen.

Reinforcement = dpro - kd

IFln.-Self m4

IFln.-Self = bFln.*hFln.3/12

IWeb. m4

IWeb.-Self =bWeb.*(kd - hFln.)3/3


IFln.-Neut = IFln.-Self+ bFln.*hFln.*dFln.2

m) 0.078

n) Moment of Inertia of Transformed Tensial Steel Area as Equivalent Concrete 0.191

o) Moment of Inertia for Crack Section under Service Limit State 0.268

2.683E+11

iv) Instantinous Deflection at Mid Span due to Applied Uniformly Distributed Dead Loads upon Interior Girder:

a) 8.853E-08

b) 2.695E+11in which; Ie<Ig

c) 4.161E+11

d) 2.683E+11

d) 8.1474E+08 N-mm

e) 2.887

f) 1474.427 mm

g) 4,127.905 kN-m4.128E+09 N-mm

h)

11 Calculation of Instantaneous Deflections at Mid Span against Individual Applied Forces :

i) Deflections due to Uniformly Distributed Loads (DL & LL) on Interior Girder :

a) 20.612 mm

b) 4.451 mm

ii) Deflections due to Concentrated Loads (DL & LL) on Interior Girder :

Moment of Inertia of Un-crack Portion of T-Girder about Neutral Axis of Crack IUn-Crack m4

Section IUn-Crack = IFln.-Neut. + IWeb.

ISteel-Trans m4

Area about Neutral Axis of Crack Section, = As-Trans.*dTen2

Icr m4

= IUn-Crack + ISteel-Trans. mm4

EcIe is Flexural Regidity of the Component as mentioned EcIe N/mm6


Ig is the Gross Moment of Inertia for Uncracked Section under Service Limit Ig mm4

State in mm4

Icr is the Moment of Inertia of the Crack Section in mm4 Icr mm4

Mcr is Crack Moment for the Section having value = fr(Ig/yt) Mcr



= hGir -kd

Ma is the Max. Moment of Component for which the Deformation Computed Ma

Since the the Calculated value of Effective Moment of Inertia, Ie < Ig the Gross Moment of Inertia, thus value

of Ie Prevailes for Calculation Deflections of T-Girder.

Deflection due to Uniformely Distributed Factored Dead Loads (DL) DInst-UDL-DL

= KUD-DL&LLMUD-DL.L2/EcIe.

Deflection due to Uniformely Distributed Factored Live Lane Loads (LL) DInst-UDL-LL

= KUD-DL&LLMUD-LL.L2/EcIe.

a) 0.000 mm

b) 2.288 mm

c) 0.502 mm

d) 0.878 mm

e) 5.725 mm

f) 6.424 mm

g) Total Instantaneous Deflection at Mid Span due to all Applied Forces 40.881 mm

12 Calculation of Instantaneous Deflections at Mid Span against Applied Forces under AASHTO-LRFD Provision :

i) Components for Calculation of Instantaneous Deflections at Mid Span against Applied Forces under Service Limit State (WSD) & Strength Limit State (USD) :

a)

b)

(Deformation of Concrete Structure by Dan Earle Branson; Chapter-3-1, Deflection of Concrete Beams And One-Way Slab, And Both Steel And Reinforced Concrete Composite Beams. Equation 3-1)

c) 0.104 value = 5/48

d) kN-m

e) L 24.400 m

f) 23,855.620 MPa

g) 0.269

2.695E+11

Deflection due to Factored Concentrated Dead Loads (DL) for 1st & 5th DInst-CDL-X-1&5.

Cross Girder = KCL-DL-1&5-X-Gir-SuppMCDL-X-1&5.L2/EcIe.

Deflection due to Factored Concentrated Dead Loads (DL) for 2nd & 4th DInst-CDL-X-2&4.

Cross Girder = KCL-DL2&4-X-Gir(X<L/2)MCDL-X-2&4.L2/EcIe.

Deflection due to Factored Concentrated Dead Loads (DL) for 3rd DInst-CDL-X-3.

Cross Girder = KCL-DL-3-X-Gir-MidMCDL-X-3.L2/EcIe.

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Front.

Front Wheel = KCL-LL-Wheel-Front(X>L/2)MCLL-W-Front.L2/EcIe.

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Mid.

Middle Wheel = KCL-LL-Wheel-Mid(X>L/2)MCLL-W-Mid.L2/EcIe.

Deflection due to Factored Concentrated Wheel Live Load (LL) for DInst-CLL-W-Rear.

Rear Wheel = KCL-LL-Wheel-Rear(X<L/2)MCLL-W-Rear.L2/EcIe.

DInst.-Total

For Calculation of Instantaneous Deflection at Mid Span, Let Consider the action of Applied Forces both Dead Loads (DL& Live Loads (LL) are Uniformly Distributed Load (UDL)

The Instantaneous Deflection on Simple Supported T-Girder Bridge on its Interior Beam at Mid Span Caused by Applied

Uniformly Distributed Loads (UDL) is Expressed by the Equation, DInst= KMaL2/EcIe. Here,

K is Deflection Coefficient for Uniformly Distributed Loads (DL & LL) having KUDL-DL&LL

Ma is Moments at Mid Span Caused by Either only Dead Load (DL) or Ma-DL/DL+LL

Dead Load (DL) + Live Load (LL)

L is the Span Length (C/C Distance between Bearings/Supports) = SL

Ec is Modulus of Elasticity of Concrete, Ec

Ie is the Effective Moment of Inertia of Girder Crack Section under Service Limit Ie m4

State Condition as Computed according to AASHTO-LRFD-5.4.2.4.. mm4

ii) Calculated Moments at Mid Span of T-Girder under Service Limit State (WSD) :

a) 2,449.720 kN-m

b) 4,127.905 kN-m

iii) Calculated Moments at Mid Span of T-Girder under Strength Limit State (USD) :

a) 3,129.494 kN-m

b) 6,863.350 kN-m

iv) Calculation of Instantaneous Deflections at Mid Span against Applied Forces under Service Limit State (WSD) :

a) 23.633 mm

b) Instantaneous Deflection Interior Beam at Mid Span due to Total Load 39.823 mm

v) Calculation of Instantaneous Deflections at Mid Span against Applied Forces under Strength Limit State (USD) :

a) 30.191 mm

b) Instantaneous Deflection Interior Beam at Mid Span due to Total Load 66.212 mm

13 Calculation of Long Term Deflections at Mid Span against Applied Forces under AASHTO-LRFD Provisions :

i) Calculated Instantaneous or Short Term Deflections :

a) Total Instantaneous Deflection at Mid Span Calculated against Individual Applied 40.881 mm

b) 23.633 mm

c) Instantaneous Deflection at Mid Span Calculated against Total Dead & Live 39.823 mm

d) 30.191 mm

Momenet at Mid Span for Dead Loads under Service Limit State (WSD) Ma-DL-WSD

Total Momenet at Mid Span (c.g.) for Dead Load (DL) + Live Load (LL) under Ma-Total-WSD

Service Limit State (WSD).

Momenet at Mid Span for Dead Loads (DL) under Strength Limit State (USD) Ma-DL-USD

Total Momenet at Mid Span (c.g.) for Dead Load (DL) + Live Load (LL) under Ma-Total-USD

Strength Limit State (USD).

Instantaneous Deflection Interior Beam at Mid Span due to Dead Load (DL) DInst-DL-WSD

= KMa-DL-WSDL2/EcIe.

DInst-Total-WSD

(DL + LL) = KMa-Total-WSDL2/EcIe.

Instantaneous Deflection Interior Beam at Mid Span due to Dead Load (DL) DInst-DL-USD

= KMa-DL-USDL2/EcIe.

DInst-Total-USD

(DL + LL) = KMa-Total-USDL2/EcIe.

DInst.-Total

Forces both Dead Load (DL) & Live Load (LL) under Provision of Service Limit State of Design (WSD).

Instantaneous Deflection at Mid Span Calculated against Total Dead Load (DL) DInst-DL-WSD

under Provision of Service Limit State of Design (WSD).

DInst-Total-WSD

Loads (DL + LL) under Provision of Service Limit State of Design (WSD).

Instantaneous Deflection at Mid Span Calculated against Total Dead Load (DL) DInst-DL-USD

under Provision of Strength Limit State of Design (USD).

e) Instantaneous Deflection at Mid Span Calculated against Total Dead & Live 66.212 mm

f) Since Instantaneous Deflection at Mid Span Calculated against Total Dead 39.823 mm

is Lowest, thus it is Acceptable one for Calculation of Long Time Deflection.

ii) Calculation of Long Term Deflections :

a) 0.269

2.695E+11

b) 3.000

c) 119.469 mm

14 Calculation of Rotation at Supports of Interior Girder due to Torsional Force Effect on Girders :

Morley; O.B.E. D.Sc & M.i.Mech.E, Chapter-X; "Torsion or Twisting; Article-112.Torsion of Shaft not Circularin Section & Equation-6 (For Rectangular Section) in Combination with Book" Concrete Bridge Practice, Analysis, Design And Economics"; by Dr. V. K. Raina, Chapter-24.5, Design Against Combined Shear & Torsion, Table-24.2; Relation between Torsional Shear Stress and Torsional Moment, & it Equation of Note

i) Phenomenon in Calculation of Rotation at Supports of Interior Girder :

a) The Bridge Girders are being considered as T-Beam having Dead Loads due to Self Weight & Superimposed Loads from Deck Slab, W/C, Sidewalk, Curb/Wheel Guard, Railing, Rail Posts etc. whereas Live Loads from Wheel Loads of Truck, Lane Loads, Pedestrian etc. Thus the Torsional Forces at Support is the Total Shearing Forces acting on that Section.

c)Tensional Forces due to Eccentric Loading of both Dead Loads & Live Loads. Dead Loads of Structure & Live LaneLoads have almost equilibrium actions. Whereas Truck Live Loads have a significant effect in these respect. Wheel

Line of Girder of each side of Interior Girders. Thus Location or Eccentricity of Torsional Force action will be consider

ii) Formula Related to Rotation at Support Position & Calculation of Rotation only for Main T-Girder :

a) q - radian (Book " Strength of Materials" by Arthur Morley; O.B.E. D.Sc & M.i.Mech.E, Chapter-X; "Torsion or Twisting; Article-112. Torsion of Shaft not Circularin Section & Equation-6 Related to Rectangular Section)

b) J 0.021

DInst-Total-USD

Loads (DL + LL) under Provision of Strength Limit State of Design (USD).

DInst-Acceptable.

& Live Loads (DL + LL) under Provision of Service Limit State of Design (WSD).

The Instantaneous Deflections are being Calculated Based on Ie, the Effective Ie m4

Moment of Inertia Computed according to Equation 5.7.3.6.2-1; Article 5.7.3.6.2. mm4

Since Instantaneous Deflection is Based on Ie, Multiplying Factor l = 3.000 l-Ie.

Calculated value of Long Term Deflectiom at Mid Span = lIe*DInst-Acceptable. DLong.

(In Computation of Rotations of Interior Girders the Provisions of Book " Strength of Materials" by Arthur

(ii)T,I or L Section Fot Torsional Costant-J)

Bridge Interior T-Girders has a Flange Width, b = 2.000m, & Web bWeb = 0.450m. Girders are being effect by the

Position of a moving Truck may change over Bridge Deck & will have effect up to a distance of b/2 from the Center

at a distance b/4 from Girder Center Line.

Roational Angle at Support, q = (40J/A4)*((MTL/2)/N); where,

J is Torsional Constant Relate to Moment of Inertia & RCC T-Girder Section.

c) A 1.030

1030000.000

d) - kN-m N-mm

e) L 24.400 m 24,400.000 mm

f) N - Mpa = Shear Stress/Shear Strain.

g) - kNthe Calculated Total Shearing Forces at Support.

h) 0.500 m

500.000 mm

i) - kN-m

j) -

k) - mm/mm

l) N -

m)

J 0.021 ("Concrete Bridge Practice, Analysis, Design And Economics"; by Dr. V. K. Raina, Chapter-24.5, Design Against Combined Shear & Torsion, Table-24.2; Relation between Torsional Shear Stress and Torsional Moment, & it Equation of Note (ii)T,I or L Section Fot Torsional Costant-J)

n)

o)

Value of y/x 1 1.5 2 3 5 >5Value of g 0.14 0.2 0.23 0.26 0.29 0.33

p) Calculated value of g for Different Rectangular Sections of T-Girder in respect of y/x Ratio is Shown in Table below :

Notation of Notation of Value of Notation of Value of Value of Value of

Long Arm Long Arm Short Arm Short Arm Ratio y/x gm m

2.000 0.350 5.714 0.290

A is Total Area of RCC T-Girder Section = bFln*hFln +(hGir - hFln)*bWeb m2

mm2

MT is Tortional Moment at Support Position of Girder MT

L is Span Length of Girder ( C/C Distance between Support Points) = SL

N is Modulus of Regidity or Shear Modulus of Elesticity of Support Section

Under any Combination of Load, Total Torsional Forces at Support, VT = VSupp. VT

Based on the assumption of Torsional Forces (DL & LL) action Positions, the eTor.

Eccentricity or Torsional Moment Arm, eTor. = b/4 from Center of Girder.

Tortional Moment MT = VT*eTor MT

Calculated Shear Stress for Support Section = vSupp vSupp N/mm2

Calculated Shear Strain for Support Section = eSupp. eSupp.

N is Modulus of Regidity or Shear Modulus of Elesticity of Support Section N/mm2

= Shear Stress/Shear Strain = vSupp/eSupp

For T-Girder Section the value of J is Expressed by the Equation

J = g1*bWeb3*hGir + g2*hFln3(bFln - bWeb)/2 + g3*hFln

3(bFln - bWeb)/2; Here,

g is a Factor dependent upon the Ratio of Longer Arsm (y) to Shorter Arm (x) as mentioned in the Table Below :

Table Showing the values of Factor g in respect of y/x.

g Factor

g1 hGir bWeb

0.825 0.200 4.125 0.290

0.825 0.200 4.125 0.290

iii) Calculation of Rotation at Support Position Under Strength Limit State (USD) for Main Girder :

a) q 2.449E-15 radian

b) J 0.021

c) A 1.030

1030000.000

d) 430.383 kN-m 397.168*10^6 N-mm

e) L 24.400 m 24,400.000 mmy

f) N 1,596.325 Mpa = Shear Stress/Shear Strain.

g) 860.766 kNthe Calculated Total Shearing Forces at Support.

h) 0.500 m

500.000 mm

i) 430.383 kN-m

j) 1.596

k) 0.001 mm/mm

l) N 1,596.325

m)

J 0.021

iv) Calculation of Rotation at Support Position Under Service Limit State (WSD) for Main Girder :


b) J 0.021

c) A 1.030

g2 (bFln-bWeb)/2 hFln

g3 (bFln-bWeb)/2 hFln




mm2

MT is Tortional Moment at Support Position under Strength Limit State (USD ) MT-USD



under Strength Limit State (USD) Total Torsional Forces at Support, VT = VSupp. VT-USD



Tortional Moment MT = VT*eTor MT-USD











1030000.000

d) 279.032 kN-m 320.596*10^6 N-mm

e) L 24.400 m 24,400.000 mmy



h) 0.500 m

500.000 mm

i) 279.032 kN-m

j) 1.596

k) 0.001 mm/mm

l) N 1,596.325

m)

J 0.021

v) Calculation of Rotation at Support Position due to Live Loads only Under Service Limit State (WSD) :


b) J 0.021

c) A 1.030

1030000.000

d) 279.032 kN-m 397.168*10^6 N-mm

e) L 24.400 m 24,400.000 mmy


mm2




Under Service Limit State the Total Torsional Forces at Support, VT = VSupp. VT














mm2





h) 0.500 m

500.000 mm

i) 279.032 kN-m

j) 1.596

k) 0.001 mm/mm

l) N 1,596.325

m)

J 1,596.325

15 Calculation of Instantaneous Deflections at Mid Span of Cross Girders against Applied Forces :

Design And Economics; by Dr. V. K. Raina, Chapter-20, Practical Structural Analysis, Table-20.5; for Moment; Shear; Deflection; Special Cases are being Followed. Necessary Modifications are being made to introduce Moment in Place of Load for Particular Cases. AASHTO-LRFD Service Limit State of Design (WSD) are being followed in Calculation Loads both for Dead & Live Loads.)

i) Dimensional Data of Cross-Girder :

a) Span Length of Cross-Girder (Clear distance between Main Girder Faces) 1.650 m

b) Thickness of Deck Slab 0.200 m

c) Thickness of Wearing Course 0.075 m

d) Number of Cross Girders 5.000 nos

e) Depth of Cross Girders (Including Slab as T-Girder) 1.900 m

f) Width of Cross-Girder Web 0.250 m

g) Width of Main-Girder Web 0.350 m

h) C/C Distance Between Main Girders 2.000 m

i) Flenge Width of Cross-Girder 0.413 m

j) C/C Distance in between Cross-Girders in Longitudinal Direction 6.250 m

k) Filets : i) X-Girder in Vertical Direction 0.075 m

ii) X-Girder in Horizontal Direction 0.075 m

ii) Computation of Daed Loads on Cross-Girders :

a) 10.335 kN/m

Under Service Limit State the Total Torsional Forces at Support, VT = VSupp. VT











(In Computation of Rotations of Interior Girders the Provisions of Book " Concrete Bridge Practice, Analysis,

SL-X-Gir.

tSlab.

tWC

NX-Gir.

hX-Gir.

bX-Web.

bMain-Web.

C/CD-Gir.

bFln-X-Gir.

CD-X-Gir.

FX-Gir-V.

FX-Gir-H.

Dead Load due Self Weight (Excluding Slab & W/C but including Fillets) in kN DLX-Gir-Self

for per Meter Span Length = wc*((hX-Gir - tSlab)*bX-Web + 2*0.5*FX-Gir-V*FX-Gir-H)

b) 1.980 kN/m

c) 0.557 kN/m

d) 12.872 kN/m

iii) Computation of Live Loads on Cross-Girders :

a) Concentrated Live Load Reaction due to Rear/Middle Single Wheel from 72.500 kN

b) 1.279 kN/m

iv) Factored Dead Load for per meter Length of Cross-Girder under Service Limit State (WSD):

a) 12.315 kN/m

b) 0.557 kN/m

c) 12.872 kN/m

vii) Factored Live Loads of Cross-Girder under Service Limit State (WSD):

a)

72.500 kN

b) 1.279 kN/m

viii) Calculation of Factored Moments at Mid Span of Cross-Girder due to Applied Loads (DL & LL) :

a) 1.650 m

a) 2.190 kM-m

b) 0.218 kM-m

c) 12.34 kM-m

Dead Load from Slab (Within Flange Width) in kN for per Meter Span Length DLX-Gir-Slab.

= gc*bFln-X-Gir.*tSlab

Dead Load from Wearing Course (Within Flange Width) in kN for per Meter DLX-Gir-WC.

Span Length = gWC*bFln-X-Gir.*tWC

Summation of Dead Loads of Cross-Girder for per meter Length åDLX-Gir.

LLX-Gir.-Wheel.

Truck in kN = LLRSW-Load or LLMSW-Load

Uniformly Distributed Live Load due to Loan Load Reaction in kN per Meter LLX-Gir-Lane.

Span Length = LLLane/3*bFln-X-Gir.

Factored Dead Load of Cross-Girder due to Self & Slab in kN/m FDLX-Gir+Slab.

= gDC*(DLX-Gir-Self + DLX-GirSlab)

Factored Dead Load of Cross-Girder due to Wearing Course in kN/m FDLX-Gir-WC

= gDW*DLX-Gir-WC

Summation of Factored DL of Cross-Girder for per meter Length åFDLX-Gir.

Factored LL of Cross-Girder due to Wheel Load in kN;

= mgLL-Truck*IM*LLX-Gir-Wheel FLLX-Gir.-Wheel.

Factored LL of Cross-Girder due to Lane Load in kN/m FLLX-Gir-Lane.

= mgLL-Lane*LLX-Gir-Lane.

Span Length of Cross-Girder (Distance between Main Girder Faces) = SL-X-Gir. LX-Gir.

Moment due to Dead Loads (DL) under Service Limit State (WSD) MX-Gir.-DL

= åFDLX-Gir.*(LX-Gir.)2/16

Moment due to Factored Live Lane Load (LL) at Middle of Cross-Girder in kN; MX-Lane-LL.

under Service Limit State (WSD) = FLLX-Gir.-Wheel.*(LX-Gir.)2/16

Moment due to Factored Wheel Load (LL) at Middle of Cross-Girder in kN; MX-Wheel-LL.

under Service Limit State (WSD) = FLLX-Gir.-Wheel.*(L-X-Gir.)2/16

d) 14.744 kM-m

viii) Events related to Flexural Design of Cross Girder against Max. Moment under Service Limit State :

a) Provided Steel Area at Mid Span against Max. (+) ve. Moment 474,737.910

b) Effective Depth of Provided Tensial Reinforcement 10.000 mm

c) 2.907 mm

d) 7.093 mm

e) Diameter of Provided Main Tensial Reinforcement Bars 20.000 mm

f) Number of Provided Main Tensial Reinforcement Bars 4.000 nos.

g) Number of Layers for Provided Main Tensial Reinforcement Bars 1.000 nos.

h) 20.000 mm 0.020 m

h) Transformed Steel Area as Equivalent Concrete Area Provided for Tensile ###

3.798

ix) Deflection Coefficient 'K' for Different Applied Loads to Compute the Deflections at Mid Span of cross Girder :

a)

b) 0.042

c) 0.042 Girder having value = 8/192

x) Calculations for Moment of Inertia of Cross T-Girder Section about its Natural Axis for Un-crack Section :

a)

0.413 m

0.200 m 0.896 m

0.796 m

0.850 m 1.900 m

0.154 m

Total Factored Moment at Middle of Cross-Girder in kN; due all Applied Loads MTotal-X-Mid.

(DL + LL )under Service Limit State (WSD) = MX-Gir.-DL+MX-Lane-LL.+ MX-Wheel-LL

As-pro. mm2

dpro.

Depth of Neutral Axis from Top of Extreme Compression Face = kdpro dComp.

Depth of Tensial Bar Centriod from Neutral Axis = dpro - kd dTen

DBar

NBar

NLayer

Total Height of 1 Layers 20f Steel Bars if placed without Spacing = NLayer*DBar hSteel

As-Trans. mm2

Reinforcement = nAs-pro m2

Since Cross-Girders have both Fixed Ends, thus value of Deflection Coefficient K accordingly.

Deflection Coefficient for Uniformly Distributed Load (DL & LL), the value KUD-DL&LL

= 16/384 (Applicable for Dead Loads of X-Girder, Slab, W-Course & Lane Live Load).

Deflection Coefficient for Concentrated Live Load (LL) at Mid Span of Cross KCL-LL-Mid

Sketch Diagram of Cross-Girder Section :

bX-Fln =

hFln = dNeut-Axis

dFln =

hX-Web/2 = hX-Gir =

dWeb =

0.250 m

b) 0.413 m

c) 0.200 m

d) 1.900 m

e) 0.250 m

f) 0.896 m

g) 0.796 m

h) 0.154 m


j) Moment of Inertia of Web Portion about its Neutral Axis, 0.102

k) 0.052

l) 0.112

m) 0.165

n) 0.165

1.650E+11

xi) Calculations for Moment of Inertia of Cross-Girder Crack Section about Natural Axis of Section :

a)

bX-Web =

Flange Width of Cross-Girder bX-Fln.

Flange Height of Cross-Girder. hFln.

Depth of Cross-Girder. hX-Gir.

Width of Cross-Girder Web bWeb.

Distance of Cross-Girder Neutral Axis from Flange Top, dNeut-Gir. dNeut-Axis

= (bX-Fal*hFln.*hFln./2+bX-Web.*(hX-Gir.-hFln.)*((hX-Gir-hFln.)/2+hFln))/

(bX-Fln.*hFln.+bX-Web*(hX-Gir-hFln.))

Distance between Neutral Axis of Cross-Girder & Neutral Axis of Flange Portion dFln.

= dNeut-Axis - hFln./2

Distance between Neutral Axis of Web Portion & Neutral Axis of Cross-Girder dWeb.

= ((hX-Gir.- hFln.)/2+ hFln.)-dNeut-Axis

IFln. m4

IFln. = bFln.*hFln.3/12

IWeb. m4

IWeb. = bX-Web.*(hX-Gir - hFln.)3/12

Moment of Inertia of Flange Portion about Cross-Girder Neutral Axis, IFln.-Neut. m4

IFln.-Neut = IFln. + bX-Fln.*hFln.*dFln.2

Moment of Inertia of Web Portion about Cross-Girder Neutral Axis, IWeb.-Neut. m4

IWeb.-Neut = IWeb. + bX-Web.*(hWeb-hFln.)*dWeb.2

Moment of Inertia of Cross-Girder about its Neutral Axis = IX-Fln.-Neut.+ IX-Web.-Neut. IX-Gir. m4

Gross Moment of Inertia for (Un-cracke) under Service Limit State = IX-Gir. Ig-X. m4

mm4

Sketch Diagram of Cross-Girder Crack Section :

0.413 m

0.200 m

-0.197 m (0.097) 0.003 m

0.010 m 0.001 m 1.897 m

1.900 m

3.798

0.250 m

b) 0.413 m

c) 0.200 m

d) Depth of T-Girder Crack Section (Distance from Extreme Compression Fiber up kd 0.003 mto Flexural Neutral Axis of the Section).

e) 0.250 m

f) Distance between Neutral Axis of Crack Section & Neutral Axis of Flange Portion (0.097) m

g) (0.197) m

h) Distance between Neutral Axis of Crack Section & Centroid of Tensile 0.007 m


j) Moment of Inertia of Uncrack Web Portion about Neutral Axis of Crack Section (0.001)

k) 0.001

m) 0.000

n) Moment of Inertia of Transformed Tensial Steel Area as Equivalent Concrete 3.798E-06

bFln =

hFln =

dWeb = dFln. = kd =

dpro = dTen = yt =

hGir =

As-Trans. = m2

bWeb =

Flange Width of Cross-Girder bFln.

Flange Height of Cross-Girder. hFln.

Width of Cross-Girder Web bWeb.

dFln.

= kd - hFln./2

Depth of Un-crack Web Portion of Cross-Girder dWeb.

= (kd.- hFln.)

dTen.

Reinforcement = dpro - kd

IFln.-Self m4

IFln.-Self = bFln.*hFln.3/12

IWeb. m4

IWeb.-Self =bWeb.*(kd - hFln.)3/3


IFln.-Neut = IFln.-Self+ bFln.*hFln.*dFln.2

Moment of Inertia of Un-crack Portion of Cross-Girder about Neutral Axis of Crack IUn-Crack m4

Section IUn-Crack = IFln.-Neut. + IWeb.

ISteel-Trans m4

o) Moment of Inertia for Crack Section under Service Limit State 0.000

4.185E+08

xii) Instantinous Deflection at Mid Span due to Applied Uniformly Distributed Dead Loads upon Cross Girder:

a) 1.282E-16

b) 8.126E+14in which; Ie>Ig

c) 1.650E+11

d) 4.185E+08

d) 2.5107E+08 N-mm

e) 2.887

f) 1897.093 mm

g) 14.744 kN-m1.474E+07 N-mm

h)

xiii) Instantinous Deflections due to Uniformly Distributed Loads (DL & LL) on Cross-Girder Mid Span :

a) 6.271E-06 mm

b) 3.556E-04 mm

c) 3.556E-04 mm

d) Total Deflection at Mid Span of Cross Girder Due to all Applied Forces. 0.001 mm

xix) Computation of Rotations at Support Position of Girder due to Different Applied Loads :

a) Since the Cross-Girders have Rigid Fixed Ends, thus Rotation on their Support 0.000 Radian.

Area about Neutral Axis of Crack Section, = As-Trans.*dTen2

Icr-X m4

= IUn-Crack + ISteel-Trans. mm4

EcIe is Flexural Regidity of the Component as mentioned EcIe N/mm6


Ig-X is the Gross Moment of Inertia for Uncracked Section under Service Limit Ig-X mm4

State in mm4

Icr is the Moment of Inertia of the Crack Section in mm4 Icr-X mm4

Mcr is Crack Moment for the Section having value = fr(Ig/yt) Mcr-X



= hGir -kd

Ma is the Max. Moment on Component for the Section = MTotal-X-Mid. Ma

Since the the Calculated value of Effective Moment of Inertia, Ie > Ig-X the Gross Moment of Inertia, thus value

of Ig-X Prevailes for Calculation Deflections of T-Girder.

Deflection due to Uniformely Distributed Factored Dead Loads (DL) DInst-X-UDL-DL

= KUD-DL&LLMX-Gir.-DL.(LX-Gir.)2/EcIg-X.

Deflection due to Uniformely Distributed Factored Live Lane Loads (LL) DInst-X-UDL-LL

= KUD-DL&LLMX-Lane-LL.(LX-Gir.)2/EcIg-X.

Deflection due to Concentraded Factored Live Wheel Loads (LL) DInst-X-CL-LL

= KCL-LL-MidMX-Wheel-LL.(LX-Gir.)2/EcIg-X.

DInst-X-Total

a-X-Gir-DL+LL

Position = 0.

13 Calculation of Deformation due to Creep & Shrinkage under Provisions of AASHTO-LRFD-5.4.2.3 :

i) Provisions of Deflections due of Creep & & Shrinkage for RCC Bridge Structures :

a) In RCC Bridge Structure Calculation of Deformation under Creep & Shrinkage is not Essential Component, rather than that are Applicable for Prestressed Concrete Bridge Structures. In Calculation of Deformation/Deflections in Prestressed

ii) Calculation of Creep Coefficient under AASHTO-LRFD-5.4.2.3-2:

a)in which,

b) 0.984

c) H 90 %

d) 0.65

V/S 150.000 mm

155.263 mm

150.000 mm

e) 29.500

f) 190.000

g) t 1825 daysto Sea Shore with Higher Intensity of Humidity, thus the Concrete will gain MaturityEarlier than other Areas. Considering the situation it is considered 5 years or 1825 days as Maturity Period of Concrete.

h) 28 daysConventionally it is Considered Concrete gains full Strength within 28 days, thus it may also taken as Initial Load Time.

i) 21.000 Mpa

j) 1.128

iii) Calculation of Shrinkage Strain under AASHTO-LRFD-5.4.2.3-3:

Concrete Bridge Structures Articles 5.9.5.3 & 5.9.5.4 in Conjunction with Article 5.7.3.6.2 are to be Considered.

The Creep Coefficient is Expressed by Y(t*ti) = 3.5kckf(1.58 - H/120)*ti-0.118*(t - ti)0.6/(10.0 +(t - ti)0.6). Equation-5.4.2.3.2-1

kf is Effect of Concrete Strength having value = 62/(42+f/c) kf

H is Relative Humidity in percentage. Since Bridge is Located on Sea Shore, Let Consider Value of H = 90%

kc is a Factor for the Effect of the Volume to Surface Area Ratio of the Component kc

as Mentioned in Figure-5.4.2.3.2-1. For (t -ts) = 1825 days, kc = 0.65 According to C-5.4.2.3.2 the Max. value of Ratio V/S is 150 mm.

i) Calculated values of VGir/SArea.

ii) Allowable Max. value of Ratio VGir/SArea. According to C-5.4.2.3.2

VGir is Volume of an Interior T-Girder = LGir*((hGir- tSlab)*bWeb + (tSlab + tWC)*bFln) VGir. m3

SArea is the Surface Area of an Interior T-Girder and Calculations are being done SArea. m2

Considering only Exposed Faces in Longitudinal Direction having value

= LGir*(2*(hGir- tSlab) + bGir-Web + (bFln - bGir-Web) +bFln)

t is the Maturity Period of Concrete in days. Since Bridge is Located very close

ti is Age of Concrete when Initial Load is allowed for the Structure in days. tf

f/c is Ultimate Compressive Strength of Concrete f/

c

Creep Coefficient Y(t*ti) = 3.5kckf(1.58 - H/120)*ti-0.118*(t - ti)0.6/(10.0 +(t - ti)0.6). Y(t*ti)

a)

b) t 28 days

c) 0.10

d) 0.86

e) -1.385

For Moist Cured Concretes devoid of Shrinkage-prone Aggregates, the Strain due Shrinkage is Express by the Equation

esk = -kskh *(t/(35.00 + t)*0.51*10-3,where,

t is Drying Time in days for Curing Concrete after Casting having value = 28days.

ks is Size Factor Specified in Figure-5.4.2.3.3-2. For Ratio V/S = 150, ks = 0.10 ks

kh is Humidity Factor Specified in Table 5.4.2.3.3-1. For 80% of Humidity the kh

value kh = 0.86

Thus Strain due to Shrinkag esk = -kskh *(t/(35.00 + t)*0.51*10-3 esk

kN/m/m-Wd.

N/mm/mm-Wd.

ii) Consideration of Deflection is Mandatory for Provision of Precast Reinforced Concrete Three-side Structure According

iii) Consideration of Deflection is Mandatory for Provision of Metal Grid Decks and other Light Weight Metal and Concrete

In Computation of Deflection the Service Limit State should be Considered in Calculation of Moments against Respective

Except for

should be Considered for Respective Live Loads in Combination with the

Bridge on its Interior Beam at Mid Span Caused by Applied ) is Expressed

) whether Uniformly Distributed Load, ) Deflection Coefficient also depends upon Position

For Calculation of Instantaneous Deflection at Mid Span, Let Consider the action of Applied Forces both Dead Loads (DL)

Bridge on its Interior Beam at Mid Span Caused by Applied

Calculation of Instantaneous Deflections at Mid Span against Applied Forces under Strength Limit State (USD) :

that are Applicable for Prestressed Concrete Bridge Structures. In Calculation of Deformation/Deflections in Prestressed

). Equation-5.4.2.3.2-1

Equation

T. Design of Beam Ledges for Abutment :




i)

9.807






ii)
















j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85




gc kg/m3

gWC kg/m3

gW-Nor. kg/m3

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.



a) 21.000 MPa

b) 8.400 MPa

c) 23,855.620 MPa

d) 2.887

e) 2.887 MPa

f) 410.000 MPa

g) 164.000 MPa

h) 200000.000 MPa


a) 8.384 n 8

b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102

2 Dimentional Data of Beam Ledges for Abutment :

i) Dimentions of Beam Ledge, Bearing Plate & other Events :

a) Length of Beam Ledge (Parallel to Traffic) 0.500 m

b) Width of Beam Ledge (Transverse to Traffic) 0.930 m

c) Depth of Beam Ledge (Under Central Girder) 0.115 mm

d) Length of Bearing Plate (Parallel to Traffic) 4.000 m

e) Width of Bearing Plate (Transverse to Traffic) 0.300 m

f) Depth of Bearing Plate 0.072 m

g) Length of Abutment Cap (Parallel to Traffic) for Beam Ledge Placing 0.700 m

h) C/C Distance (Transverse to Traffic) between Bridge Girders 2.000 m

i) Distance between Center of Bearing & the Face of Back Wall 0.325 m

j) Distance between Center of Bearing & the Face of Vertical Reinforcement 0.363 m

k) Distance between End Bearing Center & the Edge of Wing Wall C 0.675 m


c



c Ec











Value of R = 0.5*(fckj) = 0.5*(8.400*0.307*0.898) = 1.156

LB-Ladge

WB-Ladge

hB-Ladge

LB-Plate

WB-Plate

hB-Plate

LAb-Cap

SC/C-Gir.

av

af

on Open Face of Back Wall = av + CCov-Open

l) Effective Depth for Tension Reinforcement of Abutment Cap 0.538 m

m) Provided Spacing of 2-Ledge Shear Reinforcement of Abutment Cap 0.150 m

n) Cross-sectional Area of 2-Ledge Shear Reinforcement of Abutment Cap 226.195

o) Steel Area of Shear Reinforcement for per Meter Length of Abutment Cap 1,507.96

p) Effective Depth for Tension Reinforcement of Back Wall (From Open Face) 0.244 m

q) Provided Total Steel Area of Vertical Reinforcements on both Faces of Back 1,809.557

r) Applied Shear Force on Exteriod Pads due to Reactions from Exterior Girder 832.468 kN

s) Applied Shear Force on Interiod Pads due to Reactions from Interior Girder 359.356 kN

t) Applied Shear Force on Exteriod Pads due to Reactions from Exterior Girder 49.505 kN

u) Applied Shear Force on Interiod Pads due to Reactions from Interior Girder 215.333 kN

v) Steel Area of One no.Vertical Reinforcement of Back Wall on Open Face 113.097

w) Spacing of Vertical Reinforcement of Back Wall on Open Face which is the 125 mm C/C

3 Design of Beam Ledges under Provisions of Article 5.13.2.5 :

i) Sketch Diagram of Abutment Cap & Beam Ledge showing Effects/Cracks Caused by Applied Loads :

Centerline of Bearing. S C

de-Cap

spro-Cap

Av-f-Cap mm2

Avf-Cap/m mm2/m

= Avf-Cap(1.00/spro-Cap)

de-Wall

Avf-Wall mm2/m

Wall for per Meter Horizontal Length = As-Earth-V + As-Open-V

VU-Ext.-USD

under Strength Limit State (USD) = Vu-Supp-Ext.-USD

VU-Int.-USD

under Strength Limit State (USD) = Vu-Supp-int.USD

VU-Ext.-WSD

under Service Limit State (WSD) = Vu-Supp-Ext.-WSD

VU-Int.-WSD

under Service Limit State (WSD) = Vu-Supp-int.WSD

Ahr mm2

which is the Steel Area of One Ledge Hanger Reinforcement = Af-12.

sHanger

Spacing for Ledge Hanger Reinforcement = spro.-V-Open

Ahr @s af

de

aV

VU VU

NUC

2

h W

2CFor Shear

Force

For Flexural &Horizontal Force

WFor Hanger

Reinforcement

L

ii) Requirement for Design the Beam Ledges under Provisions of Article 5.13.2.5.1 :

a)

b)

c)

d)

e)

iii) Design for Shear Force (Article-5.13.2.5.2) :

a)

b) 1378877.086 Mpa

c)

c-i) 3047625.000 Mpa

c-ii) 3,990,937.50 Mpa

de

W + 4aV

W + 5af

de/2

W + 3av

de/2

de/2

According to Article 5.13.2.5.1 the Design basie of Beam Ledge will be to Resist the under Mentioned Features;

Flexural, Shear & Horizontal Forces at the Location of Crack-1.

Tension Force in the Supporting Element at the Location of Crack-2.

Punching Shear at Points of Loading at the Location of Crack-3; and

Bearing Force at Location of Crack-4.

Design of Beam Ledges for Shear Force will be in accordance with the Requirements for Shear Friction Specified in Article 5.8.4.

According to Article 5.8.2.1; the Nominal Shear Resistance of the Interface Vn-Comp.

Plane is Vn = cAcv + m(Avf-Capfy + Pc), (Equation 5.8.4.1-1), where;

Vn is Nominal Shear Resistance & its value will be the Lesser of Equations Vn £ 0.2f/cAcv (Equation 5.8.4.1.2) or

Vn £ 5.5Acv (Equation 5.8.4.1.3).

Against Equation 5.8.4.1.2 the Clculated Allowable value of Vn-1 = 0.2f/cAcv Vn-1

Against Equation 5.8.4.1.3 the Clculated Allowable value of Vn-2 = 5.5Acv Vn-2

1

3 4

c-iii) 3047625.000 Mpa

d) 725,625.000

Reinforcement from Compression Face and the Width of the Concrete Face

d-i) S 2,000.000 mm

d-ii) 2,230.000 mm

d-iii) 2*C 1,350.000 mm

d-iv) 2*C 1,350.000 mm

e) 2035.7520395

f) c 0.75 MpaFor Concrete placed against clean, hardened Concrete with intentionally

g) m 1.000 For Concrete placed against clean, hardened Concrete with intentionally

g-i) l 1.000

h) 0.000 Mpa

i) Vn-comp.<Vn-allow. SatisfiedNominal Shear Resistance Force, whether the Provision of Article have Satified or Not.

j)

iv) Design for Flexural and Horizontal Force (Article-5.13.2.5.3):

a)Friction Specified in Article 5.8.4.

b) 988968.521 Mpa

Allowable value of Vn is the Minimum of Vn-1 & Vn-2 Vn-Allow.

Acv is Area of Concrete engaged in Shear Transfer in mm. The value of Acv Acv mm2

should be Calculated by Multiplying the Effective Depth de-Cap, for the Tension

Considering the Leasser One of S, (W + 4av) & 2C as shown in the Sketch

Diagram. = de-Cap*2*C (Minimum of S, (W + 4*av) & 2*C.)

From the Sketch Diagram value of S = SC/C-Gir.

From the Sketch Diagram value of (W + 4av ) = (WB-Ladge + 4av) (W + 4av)

From the Sketch Diagram value of 2*C

Applicable value for the purpose is Minimum of S, (W + 4av) & 2*C

Avf is Area of Shear Reinforcement Crossing the Shear Plan in mm2. Avf-Cap mm2

=(Av-Cap/m*2*C)/1000

c is Choesion Factor according to Article 5.8.4.2. in Mpa.

roughened surface to an amplitude of 6.00mm. C = 0.75 Mpa

m is Friction Factor according to Article 5.8.4.2.

roughened surface to an amplitude of 6.00mm. m = 1.00*l .

For Normal Density Concrete l = 1.00

Pc is Permanent Net Compressive Force Normal to Shear Plan. If Force is Pc

Tensile Pc = 0.

Relation between Computed Shear Force Vn-Comp.& Vn-Allow. the Allowable

Since the Computed Shear Force Vn-Comp.< Vn-Allow. the Allowable Nominal Shear Resistance Force, thus Design in respect of Shear Force is OK.

Design of Beam Ledges for Flexural and Horizontal Force will be in accordance with the Requirements for Shear

According to Article 5.8.2.1; the Nominal Shear Resistance of the Interface Vn-Comp.

Plane is Vn = cAcv + m(Avf-Wallfy + Pc), (Equation 5.8.4.1-1), where;

c)

c-i) 1383480.000 Mpa

c-ii) 1,811,700.00 Mpa

c-iii) 1383480.000 Mpa

d) 329400.00

Reinforcement from Compression Face and the Width of the Concrete Face

d-i) S 2,000.000 mm

d-ii) 2,745.000 mm

d-iii) 2*C 1,350.000 mm

d-iv) 2*C 1,350.000 mm

e) 1,809.56

f) c 0.75 MpaFor Concrete placed against clean, hardened Concrete with intentionally

g) m 1.000 For Concrete placed against clean, hardened Concrete with intentionally

g-i) l 1.000

h) 0.000 Mpa

i) Vn-comp.<Vn-allow. Satisfiedthe Allowable Nominal Flexural and Horizontal Resistance Force, whether the Provision of Article have Satified or Not.

j)

Vn is Nominal Shear Resistance & its value will be the Lesser of Equations Vn £ 0.2f/cAcv (Equation 5.8.4.1.2) or

Vn £ 5.5Acv (Equation 5.8.4.1.3).

Against Equation 5.8.4.1.2 the Clculated Allowable value of Vn-1 = 0.2f/cAcv Vn-1

Against Equation 5.8.4.1.3 the Clculated Allowable value of Vn-2 = 5.5Acv Vn-2

Allowable value of Vn is the Minimum of Vn-1 & Vn-2 Vn-Allow.

Acv is Area of Concrete engaged in Shear Transfer in mm2. The value of Acv Acv mm2

should be Calculated by Multiplying the Effective Depth de-Wall, for the Tension

Considering the Leasser One of S, (W + 5af) & 2C as shown in the Sketch

Diagram. = de-Wall*2*C (Minimum of S, (W + 5*af) & 2*C.)

From the Sketch Diagram value of S = SC/C-Gir.

From the Sketch Diagram value of (W + 5af ) = (WB-Ladge + 5af) (W + 5*af)

From the Sketch Diagram value of 2*C

Applicable value for the purpose is Minimum of S, (W + 5af) & 2*C

Avf is Area of Shear Reinforcement Crossing the Shear Plan in mm2. Avf mm2/m

c is Choesion Factor according to Article 5.8.4.2. in Mpa.

roughened surface to an amplitude of 6.00mm. C = 0.75 Mpa

m is Friction Factor according to Article 5.8.4.2.

roughened surface to an amplitude of 6.00mm. m = 1.00*l .

For Normal Density Concrete l = 1.00

Pc is Permanent Net Compressive Force Normal to Shear Plan. If Force is Pc

Tensile Pc = 0.

Relation between Computed Flexural and Horizontal Force Vn-Comp.& Vn-Allow.

Since the Computed Flexural and Horizontal Force Vn-Comp.< Vn-Allow. the Allowable Nominal Flexural & Horizontal Resistance Force, thus the Design in respect of Flexural and Horizontal Force is OK.

v) Design for Punching Shear (Article-5.13.2.5.4):

a) The Truncated Pyramids assumed as failure Surfaces as shown in Sketch Diagram should not overlapped.

b)

c) 19,075.316 kN

19075.316*10^3 N

d) 1,155.743 kN

1155.743*10^3 N

e) 1,499.104

1499.104*10^3 N

f) C < S/2 Not ApplicableProvision is Applicable or Not.

g) C < S/2 ApplicableProvision is Applicable or Not.

h) (C-0.5W)<de Applicablethe Provision is Applicable or Not.

i) C < S/2 ApplicableProvision is Applicable or Not.

j) C < (C-0.5*W) Not Applicablewhether the Provision is Applicable or Not.

k)) Vu-ext.<Vn-int-ext Satified

l) Vu-int.<Vn-int-ext Satified

m) Vu-ext.<Vn-int-ext Satified

n) Vu-ext.<Vn-int-ext Satified

o)

Provisions of Article-5.13.2.5.4 in respect of Nominal Punching Shear Resistance Vn in N are;

At Interior or Exterior Pads where the End Distance C is Greater than S/2, Vn-Int&Ext.

Vn = 0.328Öf/c(W + 2L + 2de)de ; (Equation-5.13.2.5.4-1).

At Exterior Pads where the End Distance C is Less than S/2 & (C-0.5W) is Vn-Ext.

Less than de ;Vn = 0.328Öf/c(W + L + de)de ; (Equation-5.13.2.5.4-2).

At Exterior Pads where the End Distance C is Less than S/2 but (C-0.5W) Vn-Ext.

is Greater than de ;Vn = 0.328Öf/c(0.5W + L + de+C)de ; (Equation-5.13.2.5.4-3).

Relation between C & S/2 according to Equ.-5.13.2.5.4-1 and whether the

Relation between C & S/2 according to Equ.-5.13.2.5.4-2 and whether the

Relation between C & (C-0.5W) according to Equ.-5.13.2.5.4-2 and whether

Relation between C & S/2 under Equ.-5.13.2.5.4-3 and whether the

Relation between C & (C-0.5W) according to Equ.-5.13.2.5.4-3 and

Relation between Applied Shear Force on Exterior Pad VU-Supp-Ext-USD & the

Computed Nominal Punching Shear Resistance Vn-Comp-Int&Ext. according to Equ.-5.13.2.5.4-1.

Relation between Applied Shear Force on Interior Pad VU-Supp-Int-USD & the

Computed Nominal Punching Shear Resistance Vn-Comp-Int&Ext.


Computed Nominal Punching Shear Resistance Vn-Comp-Ext. according to Equ.-5.13.2.5.4-2.


Computed Nominal Punching Shear Resistance Vn-Comp-Ext. according to Equ.-5.13.2.5.4-3.

From the Calculations it is being found the Provisions of Equ.-5.13.2.5.4-2 are Fully Applicable, whereas Provision of Equ.-5.13.2.5.4-1 is not Applicable at all & Provisions of Equ.-5.13.2.5.4-3 are Partially Applicable.

p)

vi) Design of Hangur Reinforcement (Article-5.13.2.5.5):

a) The hanger Reinforcement Shall be Provided in addition to the Lesser Shear Reinforcement on either Side of theBeam Reaction being supported.

b) The Vertical Reinforcement of Open Face of Back Wall with its Spacing is the Hanger Reinforcement for a Single-Beam Ledge as shown in the Sketch Diagram.

c) 353.339 kN

d) 741.919 kN

e) 113.097

f) S 2,000.000 mm

g) s is Spacing of Hanger Reinforcement or Spacing of Vertical Reinforcement s 125 mm

h) Applied Shear Force on Interiod Pads due to Reactions from Interior Girder 359.356 kN

i) Applied Shear Force on Interiod Pads due to Reactions from Interior Girder 215.333 kN

k) Vu-int.-wsd<Vn-int-wsd

Satified(Equ.-5.13.2.5.5-1).

l) Vu-int.-usd<Vn-int-usd

Satified(Equ.-5.13.2.5.5-2).

m)Safe since Bridge Girder Loads upon Abutment Cap is Directly related with Back Wall.

In all the Cases the values of Applied Shear Forces VU < Vn, the Computed Nominal Punching Shear Resistance, thus the in respect of Punching Shear the Design is OK.

For the Service Limit State (WSD) the Nominal Shear Resistance of Hanger Vn-WSD.

Reinforcement, Vn = Ahr(0.5fy)*(W + 3av)/s; (Equ.-5.13.2.5.5-1).

For the Strength Limit State (USD) the Nominal Shear Resistance of Hanger Vn-USD.

Reinforcement, Vn = (Ahrfy*S)/s; (Equ.-5.13.2.5.5-2). Where,

Ahr is Steel Area of One Ledge Hanger Reinforcement (Steel Area of One no. Ahr mm2

Vertical Reinforcement of Back Wall on Open Face) in mm2.

S is Spacing of Bearings Placing in mm. = SC/C-Gir.

of Back Wall on Open Face in mm. = sHanger

VU-Int.-USD

under Strength Limit State (USD) = Vu-Supp-int.USD (Max. of Exe.& Int.).

VU-Int.-WSD

under Service Limit State (WSD) = Vu-Supp-int.WSD (Max. of Exe.& Int.).

Relation between Applied Shear Force on Interior Pad VU-Supp-Int-WSD & the

Computed the Nominal Shear Resistance of Hanger Reinforcement Vn-WSD.

Relation between Applied Shear Force on Interior Pad VU-Supp-Int-USD & the

Computed the Nominal Shear Resistance of Hanger Reinforcement Vn-USD.

Though the Provisions for Hangur Reinforcement under Article-5.13.2.5.5 have not Satisfied, yet the Structure is

U. Design of Footing for Approach Guard Railing Post :

1 Sketch Diagram of Approach Guard Railing Arrangement :

300mm 2000mm 2000mm 300mm

225mm

75mm Railing Post175mm Railing Bars150mm175mm Ground Level150mm175mm150mm300mm Wheel Guard

450mm

300mm Footing

225mm

300mm 300mm

225mm 825mm

225mm

825mm


Description Notation Dimensions


i)

9.807






gc kg/m3

gWC kg/m3

gW-Nor. kg/m3



ii)
















j) 0.85 (AASHTO LRFD-5.7.2..2.)

k) 0.85


a) 21.000

b) 8.400

c) 23,855.620

d) 2.887

e) 2.887

f) 410.000

g) 164.000

h) 200000.000


a) 8.384 n 8

gW-Sali. kg/m3

gs kg/m3


wc kN/m3

wWC kN/m3

wW-Nor. kN/m3

wW-Sali. kN/m3

wE kN/m3


fFlx-Rin.

fFlx-Pres.

fShear.

fSpir/Tie/Seim.

fBearig.

fStrut&Tie.

fAnc-Copm-Conc.

fAnc-Ten-Steel.

fPile-Resistanc.




c



c Ec










b) r 19.524 c) k 0.291 d) j 0.903

e) R 1.102


i) Permanent & Dead Load Multiplier Factors :








ii) Live Load Multiplier Factors :


b) 1.750


d) 1.750

e) 1.750

f) 1.750

g) 1.750

h) 1.750


Value of R = 0.5*(fckj) = 0.5*(8.400*0.307*0.898) = 1.156



gEH


gEV


gES









i) 1.000



k) 1.000






q) -

r) -

t) 1.000






















gEH


gEV





b) 1.000


d) 1.000

e) 1.000



h) 1.000

i) 1.000



k) 1.000






gES


















q) -

r) -

t) 1.000

5 Dimensional Data & Loads (DL & LL) of Approach Guard Railing Post, Railing & Footing :

i) Dimensions of Railing Post, Railing & Footing :

a) Length of Railing Post Section Parallel to Traffic 225 mm

b) Width of Railing Post Section Perpendicular to Traffic 225 mm

c) Depth of Railing Section 175 mm

d) Width of Railing Section 175 mm

f) Depth of Wheel Guard Section 300 mm

g) Width of Wheel Guard Section 300 mm

h) Number of Rails 3 nos.

i) C/C Distance between Railing Posts 2000 mm

j) Total Height of Railing Post at Bridge Face upto Top of Footing 1800 mm

k) Width of Footing Section Perpendicular to Traffic 825 mm

l) Length of Footing Section Parallel to Traffic 825 mm

m) Depth of Footing for Guard Railing Post 300 mm

n) Depth of Soil over Footing for Guard Railing Post 450 mm

ii) Dead Loads from Different Components upon Footing through Railing Post:(Considering Loads upon Highest Railing Post as & Caring Equivalent Loads from Both Side)

a) 4.410 kN

b) 2.187 kN





LPost

WPost

hRail

WRail

hW-Guard

WW-Guard

WRail

C/CPost

HPost

WFooting

LFooting

hFooting

hSoil

Dead Load From 3 nos Rails = 3*wc*hRail*WRail*C/CPost PRail-DL

Dead Load due to Railing Post Self Weight = wc*LPost*WPoat*HPost PPost-DL

c) 4.320 kN

e) 4.901 kN

f) 5.513 kN

g) 21.331 kN

iii) Application of Live Loads upon Different Components of Approach Guard Railing Post & Railings :

a) Each Rail will have to Face Uniformly Distributed Live Load acting Vertically an 0.730 N/mmHorizontal Direction with a Magnitude @ 0.73 N/mm

b) Each Rail will have to Face a Concentrated Live Load acting Vertically and 890.000 Nalso in Horizontal Direction having Magnitude of 890 N at any Point.

c) 2,350.000 N

Total Height of Railing is 1500mm or Greater. The value of this Concrented

iii) Live Loads from Different Components upon Footing through Railing Post In Vertical Direction:(Considering Loads upon Highest Railing Post as & Caring Equivalent Loads from Both Side)

a) Total Live Loads due to Uniformaly Distributed Live Loads upon Railings 4.380 kN

b) Total Live Loads due to Concentrated Live Loads upon Railings 2.670 kN

c) 7.050 kN

iv) Total Unfactored Vertical Loads (DL + LL) upon Footing 28.381 kN

6 Factored Loads (DL & LL) upon Footing under Strength Limit State (USD):

i) Factored Dead Loads (DL) upon Footing under Strength Limit State (USD):

a) 5.513 kN

b) 2.734 kN

c) 5.400 kN

e) 6.126 kN

Dead Load due to Wheel Guard Weight = wc*hW-Guard*WW-Guard*C/CPost PW-Guard-DL

Dead Load due to Self Weight of Footing = wc*LFooting*WFooting*hFooting PFooting-DL

Dead Load due to Soil upon Footing = wE*LFooting*WFooting*hSoil PSoil-DL

Total Vertical Dead Load (DL) upon Footing PFooting-Total-DL

wRail-UDL

wRail-Con

Each Railing Post will have to Face a Concentrated Live Load acting at cg PLL-H

cg. Point of Top Rail or at a Point 1500 mm above the the Sidewalk, if the

Live Load is Expressed by PLL = 890+0.73L in N, Here L is the Spacing of

Post in mm = C/CPost.

PRail-UDL-LL

= 3*wRail-UDL*C/CPost

PRail-Con-LL

= 3*wRail-Con

Total Vertical Live Load (LL) upon Footing PFooting-Total-LL

PFooting-Total-UF

Factored Dead Loads (DL) from Rails = gDC*PRail-DL FPRail-DL-USD

Factored Dead Loads (DL) from Railing Post = gDC*PPost-DL FPPost-DL-USD

Factored Dead Loads (DL) from Wheel Guard = gDC*PW-Guard-DL FPW-Guard-DL-USD

Factored Dead Loads (DL) from Weight of Footing = gDC*PFooting-DL FPFooting-DL-USD

f) 7.443 kN

g) 27.215 kN

ii) Factored Live Loads (LL) upon Footing under Strength Limit State (USD):

a) 7.665 kN

b) 4.672 kN

c) 12.337 kN

iii) 39.552 kN

7 Factored Loads (DL & LL) upon Footing under Service Limit State (WSD):

i) Factored Dead Loads (DL) upon Footing under Service Limit State (WSD):

a) 4.410 kN

b) 2.187 kN

c) 4.320 kN

e) 4.901 kN

f) 4.901 kN

g) 20.718 kN

ii) Factored Live Loads (LL) upon Footing under Service Limit State (WSD):

a) 4.380 kN

b) 2.670 kN

c) 7.050 kN

iii) 27.768 kN

Factored Dead Loads (DL) from Soil upon Footing = gEV*PFooting-DL FPSoil-DL-USD

Total Factored Dead Load (DL) upon Footing åFPFooting-DL-USD

Factored Live Loads (LL) due to Uniformaly Distributed Live Loads FPRail-UDL-LL-USD

upon Railings = mgLL-PL*PRail-UDL-LL

Factored Live Loads (LL) due to Concentrated Live Loads upon FPRail-Con-LL-USD

Railings = mgLL-PL*PRail-Con-LL

Total Factored Live Load (LL) upon Footing åFPFooting-DL-USD

Total Factored (DL + LL) upon Footing PFooting-Total-USD

= åFPFooting-DL-USD + åFPFooting-DL-USD

Factored Dead Loads (DL) from Rails = gDC*PRail-DL FPRail-DL-WSD

Factored Dead Loads (DL) from Railing Post = gDC*PPost-DL FPPost-DL-WSD

Factored Dead Loads (DL) from Wheel Guard = gDC*PW-Guard-DL FPW-Guard-DL-WSD

Factored Dead Loads (DL) from Weight of Footing = gDC*PFooting-DL FPFooting-DL-WSD

Factored Dead Loads (DL) from Soil upon Footing = gEV*PFooting-DL FPSoil-DL-WSD

Total Factored Dead Load (DL) upon Footing åFPFooting-DL-WSD

Factored Live Loads (LL) due to Uniformaly Distributed Live Loads FPRail-UDL-LL-WSD

upon Railings = mgLL-PL*PRail-UDL-LL

Factored Live Loads (LL) due to Concentrated Live Loads upon FPRail-Con-LL-WSD

Railings = mgLL-PL*PRail-Con-LL

Total Factored Live Load (LL) upon Footing åFPFooting-DL-WSD

Total Factored (DL + LL) upon Footing PFooting-Total-WSD

= åFPFooting-DL-WSD + åFPFooting-DL-WSD

8 Phenomenon for Flexural Design of Reinforcements , Upward Reaction & Moment due to Applied Load :

i) Phenomenon for Flexural Design of Reinforcements for Guard Post Footing :

a) The Footing of Guard Post will be Designed as an Inverted Cantilever One having Support at Post & its Span will bethe Length between Support Face & Outer Edge of Footing.

b) 0.300 m

ii) Calculation of Upward Reaction for per Square Meter Arae of Footing :

a) 41.698

a) 58.111

c) 40.798

iii) Caculation of Moments at Face of Guard under Different Combination of Loadings :

a) 1.548 kN-m

a) 2.157 kN-m

a) 1.515 kN-m

9 Features related to Flexural Design of Reinforcements for Guard Post Footing :

i) Clear cover for Reinforcements at Different Faces of Guard Post Footing :

a) 75.000 mm

b) 50.000 mm

c) 75.000 mm



b) c Variable

Design Span Length of Footing = (LFooting - LPost)/2 S-L

Upward Reaction due to Unfactored Total Vertical Loads (DL + LL) pFooting-UF kN/m2

= PFooting-Total-UF/(LFooting*WFooting)

Upward Reaction due to Factored Total Vertical Loads (DL + LL) pFooting-USD kN/m2

Strength Limit State (USD) = PFooting-Total-USD/(LFooting*WFooting)

Upward Reaction due to Unfactored Total Vertical Loads (DL + LL) pFooting-WSD kN/m2

Service Limit State (WSD) = PFooting-Total-WSD/(LFooting*WFooting)

Moment at Face of Guard Post due to Unfactored Loads (DL + LL) MUF

= WFooting*pFooting-UF*SL2/2

Moment at Face of Guard Post due to Factored Loads (DL + LL) under MUSD

Strengtm Limit State (USD) = WFooting*pFooting-USD*SL2/2

Moment at Face of Guard Post due to Factored Loads (DL + LL) under MWSD

Service Limit State (WSD) = WFooting*pFooting-WSD*SL2/2

Let the Clear Cover at Bottom Surface of Footing, C-Cov.Bot. = 75mm, C-Cov-Bot.

Let the Clear Cover at Top Surface of Footing, C-Cov.Top = 50mm, C-Cov-Top.

Let the Clear Cover at Vertical Surface of Footing, C-Cov.Vert.. = 75mm, C-Cov-Side.

c/de-Max.



c) Variable

Variable

Variable

410.000

Variable

Variable mm

Variable mm






Variable N-mm

Variable

0.004500

4.500/10^3

4.500*10^6

2.887

c) 12991602.095 N-mm

12.992 kN-m









Centroid in mm.




thus de = ds .














Snc = (b*hFooting.3/12)/(hFooting/2)




d) Variable N-mm

e) Variable N-mm

f) Variable N-mm

g)

Section Value of Value of Actuat Acceptable M Maximum

for Unfactored Cracking Factored Allowable FlexuralCalculationDead Load As per Moment Cracking Cracking Moment Factored Min. Moment Moment

of Moment Equation Value Moment Moment of Section Moment

Moment 5.7.3.3.2-1 M (1.33*M)kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m kN-m

At Face of 1.548 12.992 12.992 12.992 15.590 2.157 2.869 15.590 15.590

Guard Post

iv)


b) 0.016

10 Flexural Design of Reinforcement for Guard Post Footing :


a) The Calculated Flexural Moment at Fave of Guard Post is of (-) ve value. The 2.157 kN-m/m

2.157*10^6 N-mm/m

15.590 kN-m/mvalue the required Reinforcement will be on Bottom Surface of Footing. 15.590*10^6 N-mm/m

b) 15.590 kN-m/m

15.590*10^6 N-mm/m


a) 12 mm

b) 113.097

c) The provided Effective Depth for the Section with Reinforcement on Bottom 219.000 mm

d) 4.889 mm







for RCC Mu



pb

pb = b*b1*((f/c/fy)*(599.843/(599.843 + fy))),


MUSD

of Calculated Moment is Less than the Allowable Minimum Moment Mr.

Thus Mr is the Governing Moment for Provision of Reinforcement. For (-) ve Mr

Since MUSD < Mr, the Allowable Minimum Moment for the Section, thus MU


Let provide 12f Bars as Reinforcement on Bottom Surface of Footing. DBot.

X-Sectional of 12f Bars = p*DBot2/4 Af-12. mm2

de-pro.

Surface, dpro = (hFooting.-CCov-Bot. -DBot/2)


e) 195.096

f) 478.254 mm,C/C

g) 150 mm,C/C

h) 622.035


a) 0.001

b) 17.318 mm


d) Mpro>Mu OK

e) ppro<pmax OK


a) 0.450

b) c 14.72 mm

c) 0.85

d) 0.067


v) Provision of Flexural Reinforcement in Other Direction of Footing :

a) Since the Footing is Semitrical in Dimension & Shape thus it is Recommended to Provide Same Reinforcements

vi) Checking Against Critical Section for Shear Force on Footing under Article 5.13.3.6:

a)


2))(1/2))




spro = 125mm,C/C

The provided Steel Area with 16f bars having Spacing 250mm,C/C As-pro-Ab-Earth-V. mm2/m

= Af-16.b/spro



Mpro










in All Directions according to Sl No.-10-ii.

Under Vertical Loads the Footing will behave as Two-way Slab. According Article 5..8.3.2 the Critical Section is

b)

c) 216.000 the neutral axis between Resultants of the Tensile & Compressive Forces due

197.1 mm 216.0 mm

d) 858.426 kN

e)

1.000

f) 2,628.000 mm

g) 216.000 mm

h) 1,300.645 kN

i) 858.426 kN

k) Applied Shearing Force for the Footing through Rectangule Guard Section 39.552 kN

j) Vn>Vu Satisfied

e)thus the Footing does not require any Shear Reinforcement.

f)Post does not Require any Shear Reinforcement, thus Flexural Design of Reinforcement for Footings for Railing

vii) Checking for Factored Flexural Resistance under Provision of AASHTO-LRFD-5.7.3.2:

a) 48.280 N-mmwhere; 48.280*10^6 kN-m

Located at Distance dv from Face of Guard Post,

According to Article 5.13.3.6.1 for Two-way Action the Critical Section Located so that its Parimeter, bo, should be

Minimum but not closser than 0.5dv from the Face of conentrated Load. Here;




h = hFooting, the Depth Footing.Thus value 0.9*de for the Section; is 0.9*de. Whereas, value of 0.72h for the Section; 0.72h

According to Article 5.13.3.6.3 For Two-way Action for Sections without Vn

Transverse Reinforcement, the Nominal Shear Resistance Vn, is Expressed

as Vn = (0.17 + 0.33/bc)*Öf/cbodv £ 0.33Öf/

cbodv ; (Equ.-15.13.3.6.3-1) where,

bc is Ratio of Long Side to Short Side of the Rectangule through which the

the Concentrated Load or ReactionForce Transmitted.= LFooting/WFooting bc

bo is Parimiter of of the Critical Section = 2*((WPost + 2*dv) + (LPost +2*dv)) bo

dv is Effective Shesr Depyh as Mentioned in SL10.vi-c dv

Computed value of Vn = (0.17 + 0.33/bc)*Öf/cbodv ; (Equ.-15.13.3.6.3-1-RH) Vn-1

Computed value of Vn = 0.33Öf/cbodv ; (Equ.-15.13.3.6.3-1-LR) Vn-2

VU

under Strengtm Limit State (USD) = PFooting-Total-USD




Since Resisting Moment > Designed Moment, Provided Steel Ratio < Max. Steel Ratio, the Footing of Guard

Guard Post is OK.


53.644 N-mm 53.644*10^6 kN-m

f 0.90

b)



e) 2.157 kN-m

2.157*10^6 N-mm

f) Mr>Ms-v-ab-usd Satisfied


a)

Where;

b) 11.118

1.515 kN-m1.515*10^6 N-mm

622.035


c) 173.49

56.00 mmTension Bar. The Depth is Summation Bottom Clear Cover & Radius of the





y(d/s-a/2)

Mn = Asfy(ds-a/2)

Mn-V-Bot


Calculated Factored Moment M at Bottom of assumed Cantilever Beam (-)MS-V-Ab-USD

in Vertcal on Earth Face = (-)åMS-V-Ab-USD







i) M is Calculated Moment for the Section under Service Limit State MS-V-Ab-WSD







of Rectangular Section, dc = DBar/2 + CCov-Bot. Since Clear Cover on Bottom Face of



17,000.00 N/mm

Since the Structure is very close to Sea, but the Footings are of Buried

246.000

d)

e) 1,089.456 N/mm




i)

Width Parameter, thus Provisions of Tensile Reinforcements on Bottom Surface of Railing Guard Post in respect

Footing, CCov-Bot = 75mm & Bar Dia, DBar = 12f ; thus dc = (12/2 + 50)mm


Cover = 50mm.In Footing the Tension Bars is in One Layer & as per Condition





Components Category having Allowable Max. value of ZMax. = 17000N/mm








Since Developed Tensile Stress of Tension Reinforcement of Footing fs-Dev.< fsa Computed Tensile Stress;



Maximum

Mr)

V. Analysis of Earthquake Effects upon Structural Elements :

1 Relevant Earthquake Data, Coefficients & Factors:

i) Seismic Performance Zones & Zone Acceleration Coefficients :

a)

Zone Acceleration Coefficients ( A ).

b) 0.075Country. The Soil constituents are mainly Silt, Sand, Clay & Deposits of

c) 0.15 Part of the Country. The Soil constituents are mostly Stiff Cohesive & Sandy Soils with Bed-rock, Clay Shells & Compacted Sandy layers.According to

d) 0.25 having Hard Rock, Stiff Clay & Clay Rocks with Competed Sand Layers. In

ii) Site Effects, Soil Profile & Site Coefficient-S; (AASHTO-LRFD-3.10.5):

a)

(AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

b) In Soil Profile Type-I; the Soil are mainly Rock of any nature individually or 1.00 having Stiff Soiles over Rocks with constitutents of stable deposite of Sands,

c) 1.20 more than 60000mm depth over Rock with constitutents of stable deposite of

d) 1.50 depth 9000mm or more with or without intervening layers of Sands or other

In Bangladesh the total Area of Country are being Divided in to 3(three) Seismic Zones consiseding various levels of Seismicity and named as Zone-1, Zone-2 & Zone-3 respectively. These Seismic Zones are being Demarcate Stripes having Positioned Digonaliy from South-East to North-West. More over the Seismic Zones have separate

Seismic Zone-1; This Zone is Located on South & South-West areas of the A1

Organic Particles. According to AASHTO-LRFD-3.10.4-Table 3.10.4.1. TheZone Acceleration Coefficient; A £ 0.09. In Bangladesh the value of A is being considered = 0.075.

Seismic Zone-2; This Zone has covered the Hill-tract Districts & Central part A2

AASHTO-LRFD-3.10.4-Table 3.10.4.1. The Zone Acceleration Coefficient; 0.09 < A £ 0.19. In Bangladesh the value of A is being considered = 0.15.

Seismic Zone-3; This Zone is Located on North-East part of the Country A3

some Areas within this Zone there exists Organic Deposite allso. According AASHTO-LRFD-3.10.4-Table 3.10.4.1. The Zone Acceleration Coefficient; 0.19 < A £ 0.29. In Bangladesh the value of A is being considered = 0.25.

Considering Soil Characteristics different Areas have been Classified in to 4 (Four) Soil Profile Types, those arei) Soil Profile Type-I, ii) Soil Profile Type-II, iii) Soil Profile Type-III & iv) Soil Profile Type-IV. Each of theseSoil Profiles have been provided with Site Coefficient (S) based on Soil Characteristics.

S1

Gravels or Stiff Clays having less than 60000 mm depth. The Site Coefficientvalue of Soil Profile Type-I, S = 1.00. (AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

In Soil Profile Type-II; Soil are Stiff Cohesive or deep Cohesionless having S2

Sands, Gravels or Stiff Clay. Site Coefficient value of Soil Profile Type-II, S = 1.20. (AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

In Soil Profile Type-III; Soil are Soft to Medium Stiff Clays and Sands with S3

d) 2.00

iii) Elastic Seismic Response Coefficient; (AASHTO-LRFD-3.10.6.1).

a)

sec.on Nominal Unfactored mass of the Component or Structure.

AS

b)

c) For any Mode with Period of Vibration Tm > 0.4 sec.Cms for that Mode will

iv) Response Modification Factor R, for Substructure ;(AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

a) -

b)

i) For Critical Category; 1.50

ii) For Essential Category; 1.50

iii) For Other Category; 2.00

c)




d)




e)




Cohesionless Soils. Site Coefficient value of Soil Profile Type-III, S =1.50. (AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

In Soil Profile Type-III; Soil are Soft Clays or Silts having depth greater than S4

12000mm. Site Coefficient value of Soil Profile Type-IV, S =2.00. (AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-1).

The Elastic Seismic Response Coefficient for a Seismic Zone having respective Soil ProfileType is expressed

as, Csm = 1.2AS/Tm2/3 £ 2.5A; where, Csm

i) Tm = Period of Vibration of the mth Mode in Second. Value of Tm is based Tm

A = Acceleration Coefficient for the respective ZoneS = Site Coefficient for the respective Zone.

For Soil Profile III with Period of Vibration for Modes other than fundamental Csm

Mode that have Periodes, Tm < 0.3 sec. the Value of Csm shall be taken as

Csm = A(0.8+ 4.0Tm )

Csm

as Csm = 3AS/Tm4/3

R for Abutment as Substructure of Bridge; RAb-Sub.

R for Wall Type Pier Large Dimension Substructure;

RPier-Wall-Crit.

RPier-Wall-Essen.

RPier-Wall-Other.

R for Wall Type Pier Large Dimension Substructure;

RPier-Wall-Crit.

RPier-Wall-Essen.

RPier-Wall-Other.

R for Reinforced Concrete Vertical Pile bents of Foundation;

RRC-Vert.-Pile.

RPier-Vert.-Essen.

RPier-Vert.-Other.

R for Reinforced Concrete Batter/Rick Pile bents of Foundation;

RRC-Batter.-Pile.

RPier-Batter-Essen.

RPier-Batter-Other.

f)




g)




h)




v) Response Modification Factor-R for Connection(Joints/Beaing/Hinge);(AASHTO-LRFD-3.10.5.1-Table 3.10.5.1-2).

a) 0.80

b) 0.80

c) 1.00

f) 1.00

vi) Application of Seismic Loads on Structure; (AASHTO-LRFD-3.10.7.2):

a) Seismic Loads are Applicable on Structure in any Lateral Direction.

b)

c) Wall Type Piers should be considered as Column in Weak Direction.

vii) Combination of Seismic Force Effects on Structure; (AASHTO-LRFD-3.10.8.):

a) The Elastic Seismic Force has Effect upon Component of Structure in Two Principal Axis, which are Perpendicular

b)

100 %

30 %

c)

100 %

30 %

R for Steel or Composite Steel & Concrete Vertical Pile bents of Foundation;

RSteel-Vert.-Pile.

RSteel-Vert.-Essen.

RSteel-Vert.-Other.

R for Steel or Composite Steel & Concrete Batter/Rick Pile bent of Foundation;

RSteel-Batter.-Pile.

RSteel-Batter-Essen.

RSteel-Batter-Other.

R for Multiple Column bents;

RMultiple-Column

RMultiple-Column.

RMultiple-Column.

R for Connection in between Superstructure to Abutment for All Category RSupr.-Ab.

R for Connection in expansion Joints within a Span of Superstructure RSupr.Expn-1Span.

R for Column, Piers or Pile bents to Cap Beam or Superstructure RCol/Pier/Pile-to-Supr.

R for Column or Piers to Foundations RCol/Pier-to-Foun.

Factor R is Applicable to both Orthogonal Axis of Substructure.

to each other having Two Load Cases.

Load Case-1, when the Principle Force is in Y-Y Direction (Along the Alignment of Bridge);

i) The Force Effect in Y-Y Direction is 100 %, PF-Y-Y-FE-Y-Y.

ii) The Force Effect in X-X Direction is 30%. PF-Y-Y-FE-X-X.

Load Case-2, when the Principle Force is in X-X Direction (Perpendicular to the Alignment of Bridge);

i) The Force Effect in X-X Direction is 100 %, PF-X-X-FE-X-X.

ii) The Force Effect in X-X Direction is 30%. PF-X-X-FE-Y-Y.

2 Calculation of Earthquake Design Forces :

ii) Bridge Location:

a) Bridge is Located on Seismic Performance Zone-2 of Bangladesh on Cox's Bazar & Teknuf Marin Drive Road.

The Soil posses Bead-rock from a Depth 1.500m, thus the Soil Profile is Type-I.(AASHTO-LRFD-3.10.5.2).

ARMYBR~1

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